What Must Be Added to 3x - 7 to Make X Squared + 4 X - 1

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Page No 42:

Question ane:

Add the post-obit expression.
(i)  2a + b + seven ; 4a + 2b + 3
(ii) threeten + y $-$ 8; y + 4$-$sevenx
(iii) 3x ^two + 5x$-$ 4 ; 8ten$-$ twox ² + 11
(iv) $5\sqrt{\mathrm{ten}}-4\sqrt{y}+2;2\sqrt{x}+\mathrm{vii}\sqrt{y}-\mathrm{five}$

Answer:

(i)

    2a + b + 7
foura + twob + iii
---------------------
6a + 3b + 10

(ii)

      threex + y

$-$

viii

$-$

7x + y +  4
---------------------

$-$

4x + twoy

$-$

4

(three)

    310 ² + fiveten

$-$

iv

$-$

iiten ² + eightx + 11
---------------------
x ² + 13x + vii

(4)

$$\begin{gathered} 5\sqrt{x}-4\sqrt{y}+2 \\ \underline{\underline{2\sqrt{x}+\mathrm{vii}\sqrt{y}-\mathrm{five}}} \\ \mathrm{vii}\sqrt{x}+3\sqrt{y}-\mathrm{iii} \end{gathered}$$

Folio No 42:

Question 2:

Subtract the second expression from the offset:

(i) 5x ²

$-$

half dozenxy + two; threex ² + 10xy -8
(ii) m ⁱⁱ due north

$-$

8 + mn ²; 7

$-$

k ² n

$-$

mn ²
(iii) 5x ⁱⁱ + ivy ²

$-$

6y + 8; x ²

$-$

fivey ^two + 2xy + 3y

$-$

x

Answer:

$$\begin{gathered} \left(\mathrm{i}\right) \\ \mathrm{five}{x}^{\mathrm{two}}-6xy+2 \\ 3x^2+10xy-\mathrm{eight} \\ \underline{--+} \\ 2x^2-\mathrm{xvi}\mathrm{ten}y+\mathrm{ten} \end{gathered}$$ $$\begin{gathered} \left(\mathrm{two}\right) \\ {m}^{2}n-\mathrm{viii}+m{\mathrm{due\; north}}^{\mathrm{ii}} \\ -m^{\mathrm{ii}}n+7-m{\mathrm{due\; north}}^2 \\ \underline{+-+} \\ 2m^{\mathrm{ii}}\mathrm{north}-15+2m{n}^2 \end{gathered}$$ $$\begin{gathered} \left(\mathrm{iii}\right) \\ \mathrm{v}{\mathrm{ten}}^2+4y^2-\mathrm{half\; dozen}y+8 \\ {x}^2-5y^2+2xy+3y-10 \\ \underline{-+--}\underline{+} \\ 4{\mathrm{ten}}^2+9y^2-2xy-9y+\mathrm{eighteen} \end{gathered}$$

Folio No 42:

Question three:

What should be added to 5x ^two + 2xy + y ² to get 3x ² + 4xy?

Answer:

We tin can get the required expression past subtracting vx ⁱⁱ + 2xy + y ² from threex ² + 4xy every bit follow:

      3x ² + 4xy
v10 ² + 2xy + y ²

$-$ $-$ $-$

-------------------------

$-$

2x ² + 2xy

$-$

y ⁱⁱ

Page No 42:

Question 4:

What should be subtracted from 2a + 6b $-$ 5 to get $-$ 3a + 2b + 3?

Answer:

We can get the required expression by subtracting

$-$

3a + 2b + 3 from twoa + sixb

$-$

5 equally follow:

     2a + sixb

$-$

five

$-$

3a + iib + 3
+

$-$ $-$

-------------------------
5a + 4b

$-$

8

Folio No 42:

Question 5:

Subtract 410 + y + 2 from the sum of iiix $-$ 2y + seven and fivex $-$ 3y $-$ viii.

Reply:

Sum of 3x

$-$

2y + 7 and 5x

$-$

3y

$-$

viii:

    3x

$-$

2y + 7
5x

$-$

3y

$-$

8
--------------------
viiiten

$-$

fivey

$-$

1

Now, subtraction of 8x

$-$

fivey

$-$

i and 4x + y  + 2:

    viiiten

$-$

vy

$-$

1
4x + y  + 2

$-$ $-$ $-$

---------------------
4x

$-$

6y

$-$

3

Page No 42:

Question 6:

Simplify :
(i) fiveten (2x + 3y)
(ii) (2x $-$ y) (threeten + 5y)
(3) (3xy ² + 410 ²) (xy $-$ threex ²)

Answer:

$$\begin{gathered} \left(\mathrm{i}\right) \\ \mathrm{v}\mathrm{ten}\left(2x+\mathrm{three}y\right) \\ =5x\times 2x+5x\times 3y \\ =\mathrm{x}{x}^2+15\mathrm{10}y \end{gathered}$$ $$\begin{gathered} \left(\mathrm{ii}\right) \\ \left(2\mathrm{ten}-y\right)\left(\mathrm{iii}x+5y\right) \\ =2\mathrm{10}\left(3x+5y\right)-y\left(3x+5y\right) \\ =6x^2\underline{+10xy-3xy}-5y^2 \\ =6{\mathrm{ten}}^2+7xy-5y^2 \end{gathered}$$ $$\begin{gathered} \left(\mathrm{iii}\right) \\ \left(3x{y}^2+4x^2\right)\left(xy-3x^{\mathrm{ii}}\right) \\ =3\mathrm{10}{y}^2\left(\mathrm{ten}y-3x^2\right)+\mathrm{four}{\mathrm{ten}}^2\left(\mathrm{ten}y-3{\mathrm{10}}^{\mathrm{ii}}\right) \\ =3{\mathrm{ten}}^{\mathrm{two}}{y}^{\mathrm{three}}-9x^3{y}^{\mathrm{two}}+\mathrm{four}{x}^{3}y-12{\mathrm{ten}}^4 \end{gathered}$$

Page No 42:

Question 7:

Divide the outset expression by the second. Write the quotient and the residuum.
(i) a ²

$-$

b; a

$-$

b

(ii) x ⁱⁱ

$-$ $\frac{1}{4x^2};x-\frac{\mathrm{i}}{2x}$

Respond:

(i)

∴ Caliber = a + b and residuum = 0

(ii)

∴ Quotient =

$x+\frac{1}{2\mathrm{ten}}$

and residual = 0

Folio No 44:

Question ane:

Factorise the following:
1. 4x $-$ 8y

Answer:

$$\begin{gathered} 4x-\mathrm{viii}y \\ =4\left(x-2y\right) \end{gathered}$$

Folio No 44:

Answer:

$$\begin{gathered} \mathrm{v}t+25t^2 \\ =\mathrm{v}t\left(\mathrm{ane}+5t\right) \end{gathered}$$

Folio No 44:

Reply:

$$\begin{gathered} x^4{y}^{\mathrm{v}}-3x^5{y}^{\mathrm{four}} \\ ={\mathrm{ten}}^4{y}^{\mathrm{four}}\left(y-3x\right) \end{gathered}$$

Page No 44:

Question 4:

x ² + xy $-$310 $-$ 3y

Answer:

$$\begin{gathered} {\mathrm{ten}}^2+xy-\mathrm{iii}\mathrm{ten}-\mathrm{three}y \\ =x\left(\mathrm{10}+y\right)-3\left(\mathrm{10}+y\right) \\ =\left(\mathrm{ten}+y\right)\left(x-\mathrm{three}\right) \end{gathered}$$

Folio No 44:

Question 5:

6ax $-$ half dozenpast $-$ 4ay + ninebx

Reply:

$$\begin{gathered} \mathrm{half-dozen}ax-\mathrm{half\; dozen}by-4ay+\mathrm{nine}b\mathrm{ten} \\ =6a\mathrm{ten}+\mathrm{ix}bx-\mathrm{four}ay-\mathrm{six}by \\ =\mathrm{three}\mathrm{ten}\left(\mathrm{ii}a+\mathrm{iii}b\right)-2y\left(\mathrm{two}a+3b\right) \\ =\left(2a+3b\right)\left(3x-2y\right) \end{gathered}$$

Folio No 44:

Question 6:

sevenx ² $-$ 21x + iixy $-$ viy

Answer:

$$\begin{gathered} 7x^2-21x+\mathrm{two}xy-\mathrm{half-dozen}y \\ =7x\left(x-3\right)+\mathrm{two}y\left(\mathrm{10}-3\right) \\ =\left(x-\mathrm{three}\right)\left(7x+2y\right) \end{gathered}$$

Page No 44:

Question 7:

twox ⁱⁱ $-$ threexy $-$ 8xy ² + 12y ³

Answer:

$$\begin{gathered} \mathrm{ii}{\mathrm{ten}}^2-3\mathrm{10}y-\mathrm{eight}\mathrm{10}{y}^{\mathrm{two}}+12y^{\mathrm{iii}} \\ =x\left(\mathrm{ii}x-\mathrm{iii}y\right)-4y^2\left(2x-3y\right) \\ =\left(\mathrm{ii}\mathrm{10}-\mathrm{three}y\right)\left(\mathrm{ten}-\mathrm{iv}{y}^2\right) \end{gathered}$$

Folio No 44:

Answer:

$$\begin{gathered} 81{\mathrm{10}}^{\mathrm{ii}}-64y^{\mathrm{two}} \\ ={\left(9\mathrm{10}\right)}^{\mathrm{ii}}-{\left(8y\right)}^{\mathrm{two}} \\ =\left(\mathrm{nine}\mathrm{10}+\mathrm{viii}y\right)\left(9\mathrm{ten}-8y\right) \end{gathered}$$

Page No 44:

Respond:

$$\begin{gathered} 27a^{\mathrm{ii}}-75b^2 \\ =3\left(9a^2-25b^2\right) \\ =3\left[{\left(\mathrm{three}a\right)}^2-{\left(5b\right)}^2\right] \\ =3\left(\mathrm{iii}a+5b\right)\left(3a-5b\right) \end{gathered}$$

Page No 44:

Answer:

$$\begin{gathered} 3a^3-3a \\ =3a\left(a^2-1\right) \\ =\mathrm{three}a\left(a^{\mathrm{ii}}-1^2\right) \\ =3a\left(a+1\right)\left(a-1\right) \end{gathered}$$

Page No 44:

Question 11:

x ⁱⁱ $-$ y ² $-$ half dozenx $-$ 6y

Respond:

$$\begin{gathered} x^{\mathrm{two}}-y^2-6x-6y \\ =\left(\mathrm{10}+y\right)\left(x-y\right)-6\left(x+y\right) \\ =\left(x+y\right)\left(\mathrm{10}-y-6\right) \end{gathered}$$

Page No 44:

Question 12:

(a + b) (c + d) $-$ a ² + b ²

Answer:

$$\begin{gathered} \left(a+b\right)\left(c+d\right)-a^2+b^2 \\ =\left(a+b\right)\left(c+d\right)-\left(a^2-b^2\right) \\ =\left(a+b\right)\left(c+d\right)-\left(a+b\right)\left(a-b\right) \\ =\left(a+b\right)\left[\left(c+d\right)-\left(a-b\right)\right] \\ =\left(a+b\right)\left(c+d-a+b\right) \end{gathered}$$

Page No 44:

Question 13:

ten ⁱⁱ + viiix + 24y $-$ 9y ⁱⁱ

Respond:

$$\begin{gathered} x^{\mathrm{ii}}+8x+24y-\mathrm{ix}{y}^2 \\ ={\mathrm{ten}}^{\mathrm{two}}-9y^2+\mathrm{viii}x+24y \\ ={\mathrm{10}}^2-{\left(3y\right)}^{\mathrm{two}}+\mathrm{eight}\left(x+3y\right) \\ =\left(x+3y\right)\left(\mathrm{10}-\mathrm{three}y\right)+8\left(\mathrm{ten}+\mathrm{three}y\right) \\ =\left(x+\mathrm{three}y\right)\left(x-3y+8\right) \end{gathered}$$

Folio No 44:

Question 14:

a ² $-$ 12ab + 36b ² $-$ 25

Respond:

$$\begin{gathered} a^{\mathrm{ii}}-12ab+36b^{\mathrm{ii}}-25 \\ =a^2-2\times a\times 6b+{\left(6b\right)}^{\mathrm{ii}}-25 \\ ={\left(a-6b\right)}^2-5^2 \\ =\left(a-6b+\mathrm{v}\right)\left(a-6b-5\right) \end{gathered}$$

Page No 44:

Question 15:

x ² + 9y ² $-$ 25m ² $-$ 16n ² + half-dozenxy + fortymn

Answer:

$$\begin{gathered} x^{\mathrm{two}}+\mathrm{ix}{y}^2-25k^2-\mathrm{sixteen}{n}^2+6\mathrm{ten}y+40\mathrm{grand}n \\ ={\mathrm{10}}^2+9y^2+6\mathrm{ten}y-25m^{\mathrm{two}}-\mathrm{xvi}{n}^2+40kn \\ =\left[{\mathrm{10}}^{\mathrm{ii}}+6xy+9y^2\right]-\left[25g^2-\mathrm{twoscore}mn+16n^{\mathrm{two}}\right] \\ =\left[x^2+2\times x\times \mathrm{iii}y+{\left(3y\right)}^2\right]-\left[{\left(\mathrm{v}m\right)}^{\mathrm{two}}-2\times 5\mathrm{one\; thousand}\times 4n+{\left(4\mathrm{north}\right)}^{\mathrm{ii}}\right] \\ ={\left(\mathrm{10}+3y\right)}^2-{\left(5m-\mathrm{iv}n\right)}^2 \\ =\left[\left(\mathrm{ten}+\mathrm{three}y\right)+\left(\mathrm{five}\mathrm{yard}-4n\right)\right]\left[\left(x+3y\right)-\left(5m-4n\right)\right] \\ =\left(x+\mathrm{three}y+5m-4\mathrm{north}\right)\left(x+\mathrm{three}y-5m+4\mathrm{north}\right) \end{gathered}$$

Folio No 46:

Question 1:

Factorise the post-obit.
810 ³ + 125y ⁱⁱⁱ

Respond:

$$\begin{gathered} 8x^{\mathrm{iii}}+125y^{\mathrm{iii}} \\ ={\left(\mathrm{ii}\mathrm{ten}\right)}^{\mathrm{three}}+{\left(5y\right)}^3 \\ =\left(2\mathrm{ten}+5y\right)\left[{\left(\mathrm{ii}x\right)}^2-2x\times 5y+{\left(\mathrm{v}y\right)}^2\right] \\ =\left(\mathrm{two}x+\mathrm{v}y\right)\left(4x^{\mathrm{two}}-10\mathrm{10}y+25y^{\mathrm{ii}}\right) \end{gathered}$$

Folio No 46:

Reply:

$$\begin{gathered} 2a^{\mathrm{iii}}-54b^{\mathrm{iii}} \\ =2\left(a^{\mathrm{iii}}-27b^3\right) \\ =2\left[a^3-{\left(3b\right)}^3\right] \\ =2\left(a-\mathrm{iii}b\right)\left[a^{\mathrm{ii}}+a\times 3b+{\left(3b\right)}^2\right] \\ =\mathrm{two}\left(a-\mathrm{three}b\right)\left(a^2+6ab+9b^{\mathrm{two}}\right) \end{gathered}$$

Folio No 46:

Answer:

$$\begin{gathered} {\left(a+b\right)}^{\mathrm{iii}}-8 \\ ={\left(a+b\right)}^3-{\mathrm{two}}^3 \\ =\left(a+b-2\right)\left[{\left(a+b\right)}^{\mathrm{ii}}+\left(a+b\right)\times 2+{\mathrm{ii}}^{\mathrm{two}}\right] \\ =\left(a+b-\mathrm{two}\right)\left(a^{\mathrm{ii}}+2ab+b^2+2a+2b+4\right) \end{gathered}$$

Page No 46:

Answer:

$$\begin{gathered} \frac{m^{\mathrm{iii}}}{64}+\frac{{\mathrm{north}}^3}{27} \\ ={\left(\frac{\mathrm{grand}}{4}\right)}^3+{\left(\frac{n}{3}\right)}^3 \\ =\left(\frac{\mathrm{thou}}{4}+\frac{n}{3}\right)\left[{\left(\frac{m}{\mathrm{iv}}\right)}^2-\frac{\mathrm{1000}}{\mathrm{four}}\times \frac{\mathrm{north}}{3}+{\left(\frac{n}{\mathrm{iii}}\right)}^2\right] \\ =\left(\frac{m}{4}+\frac{n}{3}\right)\left[\frac{m^2}{16}-\frac{m\mathrm{northward}}{12}+\frac{n^2}{\mathrm{nine}}\right] \end{gathered}$$

Page No 46:

Answer:

$$\begin{gathered} 8y^3-\frac{125}{y^3} \\ ={\left(2y\right)}^3-{\left(\frac{5}{y}\right)}^3 \\ =\left(\mathrm{ii}y-\frac{5}{y}\right)\left[{\left(2y\right)}^{\mathrm{two}}+2y\times \frac{5}{y}+{\left({\displaystyle \frac{5}{y}}\right)}^{\mathrm{ii}}\right] \\ =\left(2y-\frac{5}{y}\right)\left(4y^2+\mathrm{x}+\frac{25}{y^{\mathrm{ii}}}\right) \end{gathered}$$

Page No 46:

Question six:

(a + b)^three $-$ (a $-$ b)³

Answer:

$$\begin{gathered} {\left(a+b\right)}^3-{\left(a-b\right)}^3 \\ =\left[\left(a+b\right)-\left(a-b\right)\right]\left[{\left(a+b\right)}^2+\left(a+b\right)\left(a-b\right)+{\left(a-b\right)}^2\right] \\ =\mathrm{two}b\left(a^2+2ab+b^2+a^2-b^2+a^{\mathrm{ii}}-2ab+b^2\right) \\ =2b\left(3a^{\mathrm{ii}}+b^2\right) \end{gathered}$$

Page No 46:

Question vii:

(2k + 3northward)³ $-$ (3thou + 2due north)³

Respond:

$$\begin{gathered} {\left(2\mathrm{thou}+3n\right)}^{\mathrm{three}}-{\left(3m+2\mathrm{north}\right)}^3 \\ =\left[\left(2m+3n\right)-\left(3g+\mathrm{ii}\mathrm{north}\right)\right]\left[{\left(\mathrm{ii}\mathrm{thousand}+3\mathrm{due\; north}\right)}^2+\left(2\mathrm{thousand}+3n\right)\left(\mathrm{three}m+2n\right)+{\left(3k+2n\right)}^{\mathrm{ii}}\right] \\ =\left(-m+n\right)\left(4m^2+12mn+9n^2+6{\mathrm{chiliad}}^2+\mathrm{four}m\mathrm{northward}+\mathrm{ix}mn+6n^2+9{\mathrm{1000}}^2+12m\mathrm{north}+4{\mathrm{due\; north}}^{\mathrm{two}}\right) \\ =\left(-\mathrm{thou}+\mathrm{north}\right)\left(19m^2+37mn+19{\mathrm{north}}^2\right) \end{gathered}$$

Folio No 46:

Question 8:

(310 + 5y)³ $-$ (2x $-$ y)³

Answer:

$$\begin{gathered} {\left(3x+5y\right)}^{\mathrm{three}}-{\left(2\mathrm{10}-y\right)}^3 \\ =\left[\left(3x+\mathrm{five}y\right)-\left(2x-y\right)\right]\left[{\left(3\mathrm{10}+5y\right)}^2+\left(3\mathrm{ten}+5y\right)\left(\mathrm{two}\mathrm{ten}-y\right)+{\left(2\mathrm{10}-y\right)}^2\right] \\ =\left(x+6y\right)\left(9x^2+25y^{\mathrm{two}}+30xy+\mathrm{six}{x}^2-3xy+10xy-\mathrm{v}{y}^2+4x^2-4\mathrm{ten}y+y^{\mathrm{ii}}\right) \\ =\left(\mathrm{10}+\mathrm{half-dozen}y\right)\left(19x^{\mathrm{two}}+33\mathrm{10}y+21y^2\right) \end{gathered}$$

Page No 46:

Question ix:

27(x $-$1)³ + y ^three

Answer:

$$\begin{gathered} 27{\left(x-1\right)}^{\mathrm{three}}+y^{\mathrm{iii}} \\ ={\left[3\left(x-1\right)\right]}^{\mathrm{three}}+y^3 \\ =\left[\mathrm{three}\left(\mathrm{ten}-\mathrm{one}\right)+y\right]\left[{\left\{\mathrm{three}\left(x-1\right)\right\}}^{\mathrm{two}}-3\left(x-1\right)\times y+y^2\right] \\ =\left(3x+y-3\right)\left[{\left(3x-3\right)}^2-3xy+\mathrm{iii}y+y^{\mathrm{ii}}\right] \\ =\left(3x+y-3\right)\left[9{\mathrm{10}}^2-18x+\mathrm{nine}-3\mathrm{ten}y+3y+y^2\right] \end{gathered}$$

Page No 46:

Answer:

$$\begin{gathered} a^{\mathrm{half-dozen}}-b^{\mathrm{half\; dozen}} \\ ={\left(a^{\mathrm{ii}}\right)}^{\mathrm{three}}-{\left(b^2\right)}^3 \\ =\left(a^2-b^{\mathrm{two}}\right)\left[{\left(a^2\right)}^2+a^{\mathrm{ii}}{b}^2+{\left(b^2\right)}^2\right] \\ =\left(a+b\right)\left(a-b\right)\left(a^4+a^2{b}^2+b^{\mathrm{four}}\right) \end{gathered}$$

Folio No 47:

Question one:

Factorise the post-obit:
2x ² + three10 $-$ 5

Answer:

$$\begin{gathered} \mathrm{two}{x}^2+3x-\mathrm{v} \\ =2x^{\mathrm{two}}+5x-2x-5\left[\because 2\times \left(-5\right)=-10 \\ \mathrm{five}\times \left(-\mathrm{two}\right)=-10 \\ \mathrm{and}5+\left(-2\right)=\mathrm{iii}\right] \\ =x\left(2\mathrm{ten}+5\right)-\mathrm{one}\left(2x+5\right) \\ =\left(2\mathrm{ten}+5\right)\left(\mathrm{ten}-1\right) \end{gathered}$$

Page No 47:

Question 2:

iiiten ^two $-$ 14ten + eight

Reply:

$$\begin{gathered} \mathrm{iii}{x}^2-14x+\mathrm{viii} \\ =3x^2-12\mathrm{ten}-2x+8\left[\because 3\times \mathrm{viii}=24 \\ \left(-12\right)\times \left(-\mathrm{ii}\right)=24 \\ \mathrm{and}\left(-12\right)+\left(-2\right)=-14\right] \\ =3\mathrm{10}\left(\mathrm{10}-4\right)-\mathrm{ii}\left(x-\mathrm{four}\right) \\ =\left(x-4\right)\left(3x-2\right) \end{gathered}$$

Page No 47:

Question 3:

half-dozenten ^two + xix $-$ 10

Answer:

$$\begin{gathered} 6x^2+\mathrm{eleven}x-10 \\ =\mathrm{vi}{x}^2+15x-4\mathrm{10}-10\left[\because 6\times \left(-\mathrm{ten}\right)=-\mathrm{threescore}, \\ 15\times \left(-4\right)=-\mathrm{lx} \\ \mathrm{and}15+\left(-4\right)=11\right] \\ =3x\left(2x+5\right)-2\left(2x+\mathrm{v}\right) \\ =\left(\mathrm{two}x+\mathrm{five}\right)\left(3\mathrm{ten}-\mathrm{ii}\right) \end{gathered}$$

Folio No 47:

Question iv:

2x ² $-$ 710 $-$ 15

Answer:

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{ii}{\mathrm{10}}^2-\mathrm{vii}\mathrm{ten}-\mathrm{fifteen} \\ =2x^{\mathrm{ii}}+3\mathrm{10}-10\mathrm{10}-15\left[\mathrm{Since},2\times \left(-15\right)=-30 \\ 3\times \left(-10\right)=-30 \\ \mathrm{And}3+\left(-\mathrm{ten}\right)=-7\right] \\ =\mathrm{ten}\left(\mathrm{two}x+3\right)-5\left(2x+\mathrm{three}\right) \\ =\left(2x+3\right)\left(x-5\right) \end{gathered}$$

Page No 47:

Question v:

ten ² + 9xy + xviiiy ²

Answer:

$$\begin{gathered} x^{\mathrm{ii}}+\mathrm{nine}xy+18y^2 \\ =x^{\mathrm{ii}}+6xy+3\mathrm{10}y+18y^2\left[\because 1\times \mathrm{eighteen}=18, \\ \mathrm{vi}\times \mathrm{iii}=\mathrm{eighteen} \\ \mathrm{and}\mathrm{six}+3=\mathrm{ix}\right] \\ =x\left(\mathrm{ten}+\mathrm{half\; dozen}y\right)+3y\left(x+6y\right) \\ =\left(x+\mathrm{six}y\right)\left(x+3y\right) \end{gathered}$$

Page No 47:

Question 6:

a ² $-$ 5ab $-$ 36b ²

Respond:

$$\begin{gathered} a^{\mathrm{two}}-5ab-36b^2 \\ =a^2+4ab-\mathrm{ix}ab-36b^{\mathrm{two}}\left[\because 1\times \left(-36\right)=-36, \\ \mathrm{iv}\times \left(-9\right)=-36 \\ \mathrm{and}4+\left(-\mathrm{ix}\right)=-5\right] \\ =a\left(a+4b\right)-9b\left(a+\mathrm{four}b\right) \\ =\left(a+4b\right)\left(a-9b\right) \end{gathered}$$

Page No 47:

Question 7:

a ² + 14ab $-$ 51b ²

Answer:

$$\begin{gathered} a^2+14ab-51b^{\mathrm{ii}} \\ =a^2+17ab-3ab-51b^2\left[\because \mathrm{i}\times \left(-51\right)=-51 \\ 17\times \left(-3\right)=-51 \\ \mathrm{and}17+\left(-3\right)=\mathrm{fourteen}\right] \\ =a\left(a+17b\right)-\mathrm{three}b\left(a+17b\right) \\ =\left(a+17b\right)\left(a-3b\right) \end{gathered}$$

Folio No 47:

Question 8:

2grand ² + 19mn + thirtyn ²

Answer:

$$\begin{gathered} 2{\mathrm{one\; thousand}}^2+19m\mathrm{northward}+30n^2 \\ =2m^2+15mn+\mathrm{four}m\mathrm{north}+30{\mathrm{northward}}^2\left[\because \mathrm{ii}\times 30=60 \\ 15\times \mathrm{iv}=60 \\ \mathrm{and}\mathrm{fifteen}+\mathrm{iv}=19\right] \\ =\mathrm{grand}\left(2\mathrm{chiliad}+\mathrm{xv}\mathrm{north}\right)+2\mathrm{northward}\left(2m+15\mathrm{northward}\right) \\ =\left(\mathrm{ii}m+\mathrm{fifteen}\mathrm{northward}\right)\left(m+\mathrm{ii}\mathrm{due\; north}\right) \end{gathered}$$

Page No 47:

Question 9:

3a ^two $-$ 11ab + 6b ⁱⁱ

Answer:

$$\begin{gathered} \mathrm{iii}{a}^2-11ab+\mathrm{half\; dozen}{b}^2 \\ =3a^2-\mathrm{ix}ab-2ab+6b^2\left[\because 3\times 6=18 \\ \left(-9\right)\times \left(-\mathrm{ii}\right)=\mathrm{xviii} \\ \mathrm{and}\left(-\mathrm{ix}\right)+\left(-2\right)=-11\right] \\ =\mathrm{three}a\left(a-3b\right)-2b\left(a-3b\right) \\ =\left(a-3b\right)\left(3a-2b\right) \end{gathered}$$

Folio No 47:

Question 10:

6x ⁱⁱ + sevenxy $-$ 13y ⁱⁱ

Reply:

$$\begin{gathered} 6x^{\mathrm{two}}-\mathrm{vii}xy-13y^2 \\ =\mathrm{vi}{x}^2+6xy-13xy-\mathrm{thirteen}{y}^2\left[\because 6\times \left(-13\right)=-78 \\ 6\times \left(-13\right)=-78 \\ \mathrm{and}6+\left(-\mathrm{xiii}\right)=-7\right] \\ =\mathrm{half-dozen}x\left(\mathrm{10}+y\right)-13y\left(x+y\right) \\ =\left(\mathrm{10}+y\right)\left(\mathrm{six}x-13y\right) \end{gathered}$$

Page No 47:

Question eleven:

$\sqrt{\mathrm{ii}}{x}^2+3x+\sqrt{\mathrm{two}}$

Answer:

$$\begin{gathered} \sqrt{2}{\mathrm{10}}^2+3\mathrm{10}+\sqrt{2} \\ =\sqrt{2}{x}^2+2x+x+\sqrt{\mathrm{two}}\left[\because \sqrt{2}\times \sqrt{\mathrm{ii}}=2, \\ 2\times 1=2 \\ \mathrm{and}2+1=3\right] \\ =\sqrt{2}x\left(x+\sqrt{2}\right)+\mathrm{ane}\left(x+\sqrt{\mathrm{ii}}\right) \\ =\left(x+\sqrt{2}\right)\left(\sqrt{\mathrm{ii}}x+1\right) \end{gathered}$$

Page No 48:

Question 1:

Factorise the post-obit:
x ^iv $-$ 810 ² y ² + 12y ^four

Answer:

$$\begin{gathered} \mathrm{Nosotros}\mathrm{take}{x}^{\mathrm{four}}-8x^{\mathrm{ii}}{y}^{\mathrm{two}}+12y^{\mathrm{four}}. \\ \mathrm{Let}{x}^{\mathrm{two}}=a\mathrm{and}{y}^2=b \\ \mathrm{i}.\mathrm{e}.,x^{\mathrm{iv}}=a^2,y^4=b^2\mathrm{and}{\mathrm{ten}}^2{y}^2=ab \\ \therefore x^{\mathrm{four}}-8x^2{y}^2+12y^4=a^2-8ab+12b^2 \\ =a^2-\mathrm{half-dozen}ab-2ab+12b^{\mathrm{two}} \\ =a\left(a-\mathrm{half-dozen}b\right)-\mathrm{two}b\left(a-\mathrm{half-dozen}b\right) \\ =\left(a-6b\right)\left(a-2b\right) \\ \mathrm{Resubstituting}\mathrm{the}\mathrm{values}\mathrm{of}a\mathrm{and}b,\mathrm{we}\mathrm{go}: \\ {\mathrm{x}}^4-\mathrm{eight}{\mathrm{ten}}^{\mathrm{two}}{\mathrm{y}}^{\mathrm{ii}}+12{\mathrm{y}}^{\mathrm{four}}=\left({\mathrm{x}}^2-6{\mathrm{y}}^2\right)\left({\mathrm{x}}^{\mathrm{two}}-2{\mathrm{y}}^2\right) \end{gathered}$$

Page No 48:

Question two:

2x ⁴ $-$ 13x ² y ² + 15y ⁴

Answer:

$$\begin{gathered} \mathrm{We}\mathrm{have}2x^4-\mathrm{xiii}{x}^2{y}^{\mathrm{ii}}+\mathrm{xv}{y}^4. \\ \mathrm{Let}{x}^2=a\mathrm{and}{y}^2=b \\ \mathrm{i}.\mathrm{e}.,{\mathrm{ten}}^{\mathrm{iv}}=a^2,y^4=b^{\mathrm{ii}}\mathrm{and}{x}^{\mathrm{two}}{y}^{\mathrm{two}}=ab \\ \therefore 2{\mathrm{ten}}^4-13{\mathrm{10}}^2{y}^2+15y^4=2a^{\mathrm{two}}-13ab+15b^2 \\ =2a^{\mathrm{two}}-10ab-3ab+\mathrm{xv}{b}^2 \\ =\mathrm{two}a\left(a-5b\right)-3b\left(a-5b\right) \\ =\left(a-5b\right)\left(2a-3b\right) \\ \mathrm{Resubstituting}\mathrm{the}\mathrm{values}\mathrm{of}a\mathrm{and}b,\mathrm{nosotros}\mathrm{become}: \\ 2{\mathrm{x}}^4-13{\mathrm{ten}}^2{\mathrm{y}}^2+15{\mathrm{y}}^{\mathrm{iv}}=\left({\mathrm{x}}^2-\mathrm{v}{\mathrm{y}}^2\right)\left(2{\mathrm{10}}^2-3{\mathrm{y}}^2\right) \end{gathered}$$

Page No 48:

Question 3:

sixa ^iv + xia ⁱⁱ2b ⁱⁱ $-$ 10b ⁴

Answer:

$$\begin{gathered} \mathrm{Nosotros}\mathrm{have}\mathrm{half-dozen}{a}^4+11a^{\mathrm{ii}}{b}^2-\mathrm{ten}{b}^{\mathrm{four}}. \\ \mathrm{Permit}{a}^2=m\mathrm{and}{b}^{\mathrm{two}}=n \\ \mathrm{i}.\mathrm{due\; east}.,a^{\mathrm{iv}}=m^2,b^4={\mathrm{northward}}^2\mathrm{and}{a}^{\mathrm{ii}}{b}^{\mathrm{ii}}=\mathrm{1000}\mathrm{due\; north} \\ \therefore \mathrm{vi}{a}^4+11a^2{b}^{\mathrm{two}}-10b^4=\mathrm{half\; dozen}{m}^{\mathrm{ii}}+11m\mathrm{north}-10n^2 \\ =\mathrm{half\; dozen}{m}^2+15m\mathrm{due\; north}-4\mathrm{one\; thousand}n-10{\mathrm{north}}^{\mathrm{ii}} \\ =3m\left(2g+\mathrm{v}\mathrm{due\; north}\right)-\mathrm{ii}\mathrm{due\; north}\left(2m+\mathrm{v}\mathrm{due\; north}\right) \\ =\left(\mathrm{ii}m+\mathrm{v}n\right)\left(\mathrm{three}\mathrm{chiliad}-\mathrm{ii}\mathrm{northward}\right) \\ \mathrm{Resubstituting}\mathrm{the}\mathrm{values}\mathrm{of}\mathrm{grand}\mathrm{and}\mathrm{north},\mathrm{we}\mathrm{get}: \\ \mathrm{half\; dozen}{a}^{\mathrm{iv}}+11a^2{b}^2-10b^4=\left(\mathrm{two}{a}^2+\mathrm{five}{b}^2\right)\left(3a^2-2b^2\right) \end{gathered}$$

Page No 48:

Question iv:

3(x ² $-$vten)ⁱⁱ $-$ ii(x ² $-$5x + 5) $-$ half-dozen

Respond:

$$\begin{gathered} \mathrm{We}\mathrm{take}\mathrm{iii}{\left(x^2-5x\right)}^2-2\left({\mathrm{10}}^2-5\mathrm{ten}+\mathrm{five}\right)-6. \\ \mathrm{Let}{\mathrm{ten}}^2-5x=a \\ \therefore 3{\left(x^{\mathrm{ii}}-5x\right)}^2-\mathrm{ii}\left({\mathrm{ten}}^2-5x+5\right)-\mathrm{half\; dozen}=\mathrm{iii}{a}^2-\mathrm{ii}\left(a+5\right)-6 \\ =3a^{\mathrm{two}}-2a-16 \\ =3a^2-\mathrm{viii}a+6a-16 \\ =a\left(\mathrm{iii}a-8\right)+2\left(3a-\mathrm{viii}\right) \\ =\left(\mathrm{three}a-\mathrm{eight}\right)\left(a+2\right) \\ \mathrm{Resubstituting}\mathrm{the}\mathrm{value}\mathrm{of}a,\mathrm{we}\mathrm{become}: \\ \mathrm{iii}{\left(x^2-\mathrm{v}x\right)}^{\mathrm{ii}}-\mathrm{two}\left(x^2-\mathrm{v}\mathrm{ten}+5\right)-6=\left[3\left(x^{\mathrm{ii}}-\mathrm{v}x\right)-8\right]\left[\left({\mathrm{ten}}^2-\mathrm{five}\mathrm{10}\right)+2\right] \\ =\left(3{\mathrm{ten}}^2-15\mathrm{ten}-8\right)\left(x^{\mathrm{ii}}-5x+2\right) \end{gathered}$$

Page No 48:

Question 5:

(y ⁱⁱ + 5y) (y ² + 5y $-$ 2) $-$ 24

Answer:

$$\begin{gathered} \mathrm{Nosotros}\mathrm{accept}\left(y^{\mathrm{two}}+\mathrm{v}y\right)\left(y^{\mathrm{two}}+5y-2\right)-24. \\ \mathrm{Permit}{y}^2+5y=a \\ \therefore \left(y^2+\mathrm{v}y\right)\left(y^2+5y-\mathrm{ii}\right)-24=a\left(a-2\right)-24 \\ =a^2-2a-24 \\ =a^2-\mathrm{vi}a+\mathrm{four}a-24 \\ =a\left(a-\mathrm{six}\right)+4\left(a-\mathrm{half\; dozen}\right) \\ =\left(a-6\right)\left(a+\mathrm{four}\right) \\ \mathrm{Resubstituting}\mathrm{the}\mathrm{value}\mathrm{of}a,\mathrm{we}\mathrm{get}: \\ \left(y^{\mathrm{ii}}+5y\right)\left(y^2+5y-2\right)-24=\left(y^{\mathrm{ii}}+5y-\mathrm{half\; dozen}\right)\left(y^{\mathrm{ii}}+5y+\mathrm{iv}\right) \\ =\left(y^2+6y-y-\mathrm{six}\right)\left(y^2+4y+y+4\right) \\ =\left[y\left(y+6\right)-\mathrm{ane}\left(y+\mathrm{half\; dozen}\right)\right]\left[y\left(y+4\right)+1\left(y+4\right)\right] \\ =\left(y+\mathrm{six}\right)\left(y-\mathrm{ane}\right)\left(y+4\right)\left(y+1\right) \end{gathered}$$

Folio No 51:

Question 1:

Factorise the post-obit:
x ³ $-$ 27y ³ + 125 + 45xy

Answer:

$$\begin{gathered} {\mathrm{ten}}^3-27y^3+125+45xy \\ ={\left(x\right)}^3+{\left(-3y\right)}^3+{\left(5\right)}^3-3\times \left(x\right)\left(-3y\right)\left(5\right) \\ =\left[\mathrm{10}+\left(-3y\right)+5\right]\left[x^2+{\left(-\mathrm{iii}y\right)}^2+5^2-\mathrm{10}\times \left(-3y\right)-\left(-\mathrm{iii}y\right)\times 5-5\times \mathrm{ten}\right] \\ =\left(x-\mathrm{three}y+5\right)\left(x^{\mathrm{two}}+9y^2+25+3xy+15y-5\mathrm{10}\right) \end{gathered}$$

Folio No 51:

Question ii:

a ^three $-$ b ³ + viiic ³ + 6abc

Reply:

$$\begin{gathered} a^{\mathrm{iii}}-b^{\mathrm{three}}+8c^3+6abc \\ ={\left(a\right)}^3+{\left(-b\right)}^3+{\left(\mathrm{two}c\right)}^{\mathrm{three}}-3\times \left(a\right)\times \left(-b\right)\times \left(2c\right) \\ =\left[a+\left(-b\right)+2c\right]\left[a^2+{\left(-b\right)}^{\mathrm{two}}+{\left(\mathrm{two}c\right)}^{\mathrm{two}}-a\times \left(-b\right)-\left(-b\right)\times \left(\mathrm{ii}c\right)-\left(2c\right)\times a\right] \\ =\left(a-b+\mathrm{two}c\right)\left(a^2+b^{\mathrm{ii}}+4c^{\mathrm{ii}}+ab+\mathrm{two}bc-2ca\right) \end{gathered}$$

Page No 51:

Question 3:

8a ³ + 27b+ 64c ⁱⁱⁱ$-$ 72abc

Answer:

$$\begin{gathered} 8a^{\mathrm{three}}+27b^{\mathrm{iii}}+64c^3-72abc \\ ={\left(2a\right)}^{\mathrm{three}}+{\left(3b\right)}^{\mathrm{iii}}+{\left(\mathrm{iv}c\right)}^3-3\times \left(\mathrm{two}a\right)\times \left(3b\right)\times \left(\mathrm{four}c\right) \\ =\left(\mathrm{ii}a+3b+4c\right)\left[{\left(2a\right)}^{\mathrm{ii}}+{\left(\mathrm{iii}b\right)}^2+{\left(\mathrm{four}c\right)}^2-\left(2a\right)\times \left(3b\right)-\left(3b\right)\times \left(4c\right)-\left(4c\right)\times \left(\mathrm{two}a\right)\right] \\ =\left(2a+3b+4c\right)\left(\mathrm{iv}{a}^2+9b^2+16c^2-6ab-12bc-8ca\right) \end{gathered}$$

Folio No 51:

Question 4:

$-$27x ^three + y ³ $-$ z ^three $-$ ninexyz

Respond:

$$\begin{gathered} -27{\mathrm{10}}^3+y^3-z^3-\mathrm{ix}xyz \\ ={\left(-3x\right)}^3+{\left(y\right)}^3+{\left(-z\right)}^3-3\times \left(-3x\right)\left(y\right)\left(-z\right) \\ =\left[\left(-\mathrm{three}\mathrm{ten}\right)+\left(y\right)+\left(-z\right)\right]\left[{\left(-\mathrm{iii}\mathrm{ten}\right)}^{\mathrm{ii}}+{\left(y\right)}^{\mathrm{ii}}+{\left(-z\right)}^2-\left(-\mathrm{three}x\right)\times \left(y\right)-\left(y\right)\times \left(-z\right)-\left(-z\right)\times \left(-\mathrm{iii}x\right)\right] \\ =\left(-\mathrm{iii}x+y-z\right)\left(9x^2+y^{\mathrm{two}}+z^2+3\mathrm{10}y+yz-\mathrm{three}zx\right) \end{gathered}$$

Page No 51:

Question 5:

y ⁶ + 32y ³ $-$ 64

Respond:

$$\begin{gathered} \mathrm{Nosotros}\mathrm{have}{y}^6+32y^3-64. \\ \mathrm{Here},32y^3\mathrm{can}\mathrm{non}\mathrm{be}\mathrm{written}\mathrm{in}\mathrm{cubic}\mathrm{from},\mathrm{and\; then}\mathrm{will}\mathrm{split}32y^3\mathrm{into}8y^3+24y^3. \\ {y}^6+\mathrm{eight}{y}^3+24y^{\mathrm{iii}}-64 \\ =y^{\mathrm{half-dozen}}+8y^{\mathrm{iii}}-64+24y\mathrm{three} \\ ={\left(y^2\right)}^3+{\left(2y\right)}^3+{\left(-4\right)}^3-3\times \left(y^2\right)\times \left(2y\right)\times \left(-4\right) \\ =\left[y^{\mathrm{ii}}+\mathrm{two}y+\left(-\mathrm{four}\right)\right]\left[{\left(y^{\mathrm{two}}\right)}^2+{\left(2y\right)}^2+{\left(-4\right)}^2-y^2\times \mathrm{two}y-\mathrm{ii}y\times \left(-4\right)-\left(-\mathrm{iv}\right)\times y^2\right] \\ =\left(y^{\mathrm{two}}+2y-4\right)\left(y^4+\mathrm{four}{y}^2+16-2y^{\mathrm{iii}}+8y+4y^{\mathrm{ii}}\right) \\ =\left(y^{\mathrm{two}}+2y-4\right)\left(y^4-2y^{\mathrm{iii}}+\mathrm{viii}{y}^2+8y+16\right) \end{gathered}$$

Folio No 51:

Question half dozen:

x ^{half dozen} $-$ 1010 ³ $-$ 27

Reply:

$$\begin{gathered} \mathrm{Nosotros}\mathrm{have}{x}^{\mathrm{half\; dozen}}-10x^{\mathrm{three}}-27. \\ \mathrm{Here},-\mathrm{x}{x}^3\mathrm{can}\mathrm{non}\mathrm{exist}\mathrm{written}\mathrm{in}\mathrm{cubic}\mathrm{from},\mathrm{so}\mathrm{we}\mathrm{wil}\mathrm{split}-10x^3\mathrm{into}-9x^3-x^{\mathrm{three}}. \\ {x}^6-9x^{\mathrm{three}}-x^3-27 \\ =x^{\mathrm{vi}}-x^3-27-\mathrm{nine}{x}^{\mathrm{three}} \\ ={\left(x^2\right)}^3+{\left(-x\right)}^3+{\left(-3\right)}^3-3\times \left(x^{\mathrm{ii}}\right)\times \left(-x\right)\times \left(-\mathrm{iii}\right) \\ =\left[x^2+\left(-\mathrm{10}\right)+\left(-3\right)\right]\left[{\left({\mathrm{ten}}^{\mathrm{ii}}\right)}^2+{\left(-\mathrm{ten}\right)}^{\mathrm{ii}}+{\left(-3\right)}^{\mathrm{two}}-{\mathrm{10}}^{\mathrm{ii}}\times \left(-x\right)-\left(-x\right)\times \left(-3\right)-\left(-3\right)\times {\mathrm{ten}}^2\right] \\ =\left(x^2-x-3\right)\left(x^{\mathrm{iv}}+x^2+\mathrm{ix}+x^3-3\mathrm{10}+3x^{\mathrm{two}}\right) \\ =\left(x^{\mathrm{ii}}-x-3\right)\left(x^{\mathrm{iv}}+x^{\mathrm{three}}+\mathrm{four}{\mathrm{10}}^2-\mathrm{three}x+9\right) \end{gathered}$$

Folio No 51:

Reply:

$$\begin{gathered} \mathrm{We}\mathrm{have}{a}^{\mathrm{iii}}+\mathrm{iv}-\frac{1}{a^{\mathrm{three}}}. \\ \mathrm{Here},\mathrm{four}\mathrm{cannot}\mathrm{be}\mathrm{written}\mathrm{in}\mathrm{cubic}\mathrm{from},\mathrm{so}\mathrm{we}\mathrm{volition}\mathrm{split}4\mathrm{into}1+\mathrm{iii}. \\ {a}^{\mathrm{three}}+1+\mathrm{three}-\frac{\mathrm{one}}{a^3} \\ =a^3+\mathrm{ane}-\frac{1}{a^3}+\mathrm{iii} \\ =a^3+{\mathrm{i}}^{\mathrm{iii}}+{\left(-\frac{1}{a}\right)}^3-3\times a\times 1\times \left(-\frac{\mathrm{one}}{a}\right) \\ =\left[a+1+\left(-{\displaystyle \frac{1}{a}}\right)\right]\left[a^2+{\mathrm{one}}^{\mathrm{two}}+{\left(-{\displaystyle \frac{1}{a}}\right)}^2-a\times 1-1\times \left(-\frac{\mathrm{one}}{a}\right)-\left(-\frac{\mathrm{one}}{a}\right)\times a\right] \\ =\left(a+1-\frac{\mathrm{ane}}{a}\right)\left(a^{\mathrm{ii}}+1+\frac{1}{a^{\mathrm{two}}}-a+\frac{1}{a}+\mathrm{ane}\right) \\ =\left(a+1-\frac{1}{a}\right)\left(a^2+\mathrm{ii}+\frac{1}{a^2}-a+\frac{1}{a}\right) \end{gathered}$$

Page No 51:

Question eight:

(p $-$ 3q)^three + (3q $-$ 7r)³ + (7r $-$ 9)³

Answer:

$$\begin{gathered} \mathrm{We}\mathrm{accept}{\left(p-3q\right)}^3+{\left(\mathrm{iii}q-\mathrm{seven}r\right)}^3+{\left(\mathrm{seven}r-p\right)}^3 \\ \mathrm{Allow}p-3q=a,3q-7r=b\mathrm{and}7r-p=c \\ \therefore {\left(p-3q\right)}^3+{\left(\mathrm{three}q-\mathrm{seven}r\right)}^3+{\left(7r-p\right)}^{\mathrm{three}}=a^{\mathrm{iii}}+b^3+c^3 \\ \mathrm{Here},a+b+c=p-3q+\mathrm{three}q-7r+\mathrm{seven}r-p=0 \\ \mathrm{However},\mathrm{nosotros}\mathrm{know}\mathrm{that}\mathrm{if}a+b+c=0,\mathrm{then}{a}^{\mathrm{three}}+b^3+c^3=3abc \\ \therefore {\left(p-\mathrm{three}q\right)}^3+{\left(3q-\mathrm{seven}r\right)}^{\mathrm{iii}}+{\left(7r-p\right)}^3=\mathrm{three}\left(p-3q\right)\left(3q-7r\right)\left(7r-p\right) \end{gathered}$$

Page No 51:

Question nine:

(fiveten $-$ 6y)³ + (7z $-$ 5x)³ + (6y $-$ sevenz)³

Answer:

$$\begin{gathered} \mathrm{Nosotros}\mathrm{take}{\left(5x-\mathrm{six}y\right)}^3+{\left(\mathrm{vii}z-5x\right)}^3+{\left(6y-7z\right)}^{\mathrm{three}} \\ \mathrm{Let}5\mathrm{ten}-6y=a,\mathrm{vii}z-5\mathrm{ten}=b\mathrm{and}6y-7z=c \\ \therefore {\left(5\mathrm{10}-6y\right)}^3+{\left(\mathrm{seven}z-5x\right)}^{\mathrm{three}}+{\left(6y-7z\right)}^3=a^3+b^3+c^3 \\ \mathrm{Hither},a+b+c=5x-\mathrm{half-dozen}y+\mathrm{vii}z-\mathrm{five}\mathrm{ten}+6y-\mathrm{vii}z=0 \\ \mathrm{Nonetheless},\mathrm{we}\mathrm{know}\mathrm{that}\mathrm{if}a+b+c=0,\mathrm{then}{a}^{\mathrm{three}}+b^{\mathrm{three}}+c^3=3abc \\ \therefore {\left(\mathrm{v}x-6y\right)}^{\mathrm{iii}}+{\left(7z-\mathrm{v}x\right)}^3+{\left(\mathrm{half-dozen}y-\mathrm{vii}z\right)}^{\mathrm{iii}}=3\left(5\mathrm{10}-\mathrm{half-dozen}y\right)\left(7z-\mathrm{five}x\right)\left(6y-7z\right) \end{gathered}$$

Page No 51:

Question 10:

27(a $-$ b)ⁱⁱⁱ + (iia $-$ b)³ + (4b $-$ va)³

Answer:

$$\begin{gathered} \mathrm{We}\mathrm{take}27{\left(a-b\right)}^{\mathrm{iii}}+{\left(\mathrm{two}a-b\right)}^3+{\left(4b-5a\right)}^3 \\ ={\left[3\left(a-b\right)\right]}^3+{\left(\mathrm{two}a-b\right)}^3+{\left(4b-5a\right)}^{\mathrm{iii}} \\ \mathrm{Let}3\left(a-b\right)=\mathrm{10},2a-b=y\mathrm{and}4b-5a=z \\ \therefore {\left[3\left(a-b\right)\right]}^3+{\left(\mathrm{ii}a-b\right)}^{\mathrm{iii}}+{\left(4b-5a\right)}^3=x^3+y^3+z^3 \\ \mathrm{Here},\mathrm{ten}+y+z=3\left(a-b\right)+\left(2a-b\right)+\left(4b-\mathrm{five}a\right)=0 \\ \mathrm{However},\mathrm{nosotros}\mathrm{know}\mathrm{that}\mathrm{if}\mathrm{10}+y+z=0,\mathrm{and\; then}{x}^{\mathrm{three}}+y^3+z^3=\mathrm{three}\mathrm{10}yz \\ \therefore {\left[\mathrm{iii}\left(a-b\right)\right]}^{\mathrm{three}}+{\left(2a-b\right)}^3+{\left(\mathrm{iv}b-5a\right)}^{\mathrm{three}}=3\left[3\left(a-b\right)\right]\left(2a-b\right)\left(4b-\mathrm{five}a\right) \end{gathered}$$

Page No 52:

Question 1:

Which of the post-obit expressions are polynomials?
(i) five10 ²

$-$

seven10 + 4

(ii) 24

(iii) 5a ² +

$\sqrt{a}$

+ four

(iv)

$\frac{7}{8}\mathrm{10}-\mathrm{four}{x}^4+\mathrm{five}{x}^3$

(v)

$t^{\mathrm{ii}}+\sqrt{2}t-\sqrt{\mathrm{vii}}$

(6) 310 ² +

$\frac{\mathrm{seven}}{x}$ $-$

sevenx

(vii) x ³⁸

$-$

4

(viii) 3ten ^-two + x ^-one + five

(ix)

$y^{\frac{2}{3}}+7y-8$

(ten)

$\mathrm{five}\sqrt{x^3}-\mathrm{ane}$

Respond:

$\left(\mathrm{i}\right)5{\mathrm{10}}^2-7x+4\mathrm{is}\mathrm{polynomial}\mathrm{expression},\mathrm{as}\mathrm{the}\mathrm{powers}\mathrm{of}\mathrm{variable}x\mathrm{are}\mathrm{positive}\mathrm{integers}.$ $\left(\mathrm{ii}\right)24\mathrm{is}\mathrm{a}\mathrm{nonzero}\mathrm{constant}\mathrm{and}\mathrm{every}\mathrm{nonzero}\mathrm{constant}\mathrm{is}\mathrm{a}\mathrm{polynomial}\mathrm{expression}.\mathrm{Therefore},\mathrm{it}\mathrm{is}\mathrm{a}\mathrm{polynomial}\mathrm{expression}.$ $$\begin{gathered} \left(\mathrm{three}\right)\mathrm{We}\mathrm{have}5a^2+\sqrt{a}+4=\mathrm{five}{a}^2+a^{\raisebox{1ex}{$\mathrm{i}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{two}$}\right.}+4 \\ \mathrm{v}{a}^2+\sqrt{a}+\mathrm{iv}\mathrm{is}\mathrm{not}\mathrm{a}\mathrm{polynomial}\mathrm{expression},\mathrm{as}1\mathrm{of}\mathrm{the}\mathrm{powers}\mathrm{of}\mathrm{variable}a\mathrm{is}\frac{1}{2},\mathrm{which}\mathrm{is}\mathrm{not}\mathrm{an}\mathrm{integer}. \end{gathered}$$ $\left(\mathrm{iv}\right)\frac{\mathrm{seven}}{8}\mathrm{10}-\mathrm{iv}{x}^4+5x^{\mathrm{three}}\mathrm{is}\mathrm{a}\mathrm{polynomial}\mathrm{expression},\mathrm{as}\mathrm{all}\mathrm{the}\mathrm{powers}\mathrm{of}\mathrm{variable}x\mathrm{are}\mathrm{positive}\mathrm{integers}.$ $\left(\mathrm{v}\right)t^2+\sqrt{2}t-\sqrt{\mathrm{vii}}\mathrm{is}\mathrm{a}\mathrm{polynomial}\mathrm{expression},\mathrm{as}\mathrm{all}\mathrm{the}\mathrm{powers}\mathrm{of}\mathrm{variable}t\mathrm{are}\mathrm{positive}\mathrm{integers}.$ $$\begin{gathered} \left(\mathrm{six}\right)\mathrm{We}\mathrm{have}\mathrm{iii}{\mathrm{ten}}^2+\frac{\mathrm{vii}}{x}-7x=3x^2+7x^{-1}-\mathrm{vii}x \\ 3x^{\mathrm{ii}}+\frac{7}{x}-7x\mathrm{is}\mathrm{not}\mathrm{a}\mathrm{polynomial}\mathrm{expression},\mathrm{as}\mathrm{one}\mathrm{of}\mathrm{the}\mathrm{powers}\mathrm{of}\mathrm{variable}x\mathrm{is}-\mathrm{ane},\mathrm{which}\mathrm{is}\mathrm{not}\mathrm{a}\mathrm{positive}\mathrm{integer}. \end{gathered}$$ $\left(\mathrm{vii}\right){\mathrm{ten}}^{38}-4\mathrm{is}\mathrm{a}\mathrm{polynomial}\mathrm{expression},\mathrm{as}\mathrm{all}\mathrm{the}\mathrm{powers}\mathrm{of}\mathrm{variable}x\mathrm{is}\mathrm{a}\mathrm{positive}\mathrm{integer}.$ $\left(\mathrm{viii}\right)3{\mathrm{ten}}^{-\mathrm{two}}+x^{-1}+5\mathrm{is}\mathrm{not}\mathrm{a}\mathrm{polynomial}\mathrm{expression},\mathrm{as}\mathrm{the}\mathrm{powers}\mathrm{of}\mathrm{variable}\mathrm{10}\mathrm{are}-2\mathrm{and}-\mathrm{one},\mathrm{which}\mathrm{are}\mathrm{not}\mathrm{positive}\mathrm{integer}\mathrm{due\; south}.$ $\left(\mathrm{ix}\right)y^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+\mathrm{seven}y-\mathrm{eight}\mathrm{is}\mathrm{not}\mathrm{a}\mathrm{polynomial}\mathrm{expression},\mathrm{as}\mathrm{one}\mathrm{of}\mathrm{the}\mathrm{powers}\mathrm{of}\mathrm{variable}y\mathrm{is}\frac{2}{\mathrm{three}},\mathrm{which}\mathrm{is}\mathrm{not}\mathrm{an}\mathrm{integer}.$ $$\begin{gathered} \left(\mathrm{x}\right)\mathrm{We}\mathrm{have}\sqrt[5]{x^3}-1={\mathrm{ten}}^{\raisebox{1ex}{$\mathrm{iii}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{v}$}\right.}-\mathrm{ane} \\ \sqrt[\mathrm{v}]{{\mathrm{ten}}^3}-1\mathrm{is}\mathrm{not}\mathrm{a}\mathrm{polynomial}\mathrm{expression},\mathrm{as}\mathrm{the}\mathrm{ability}\mathrm{of}\mathrm{variable}x\mathrm{is}\frac{3}{5},\mathrm{which}\mathrm{is}\mathrm{not}\mathrm{an}\mathrm{integer}. \end{gathered}$$

Page No 52:

Question iii:

Write the degree of each of the post-obit polynomials:

(i) 5

(2)

$-\frac{3}{2}$

(iii)

$4p-\sqrt{\mathrm{five}}$

(iv) 3y + 4y ²

(v) x ⁹

$-$

x ⁴ +_ 10 ¹² + x

$-$

2

(vi) 5m ² n

$-$

iii

(vii) 2xy + ivx ³

(eight) sevenp ² q ⁱⁱⁱ t

$-$

11p ⁴ t + 2p ⁸

(nine) 5x ² yz³ + xy ^four z ²

(10) ab ^two cd

$-$

3a ⁱⁱ bcd ³ + 2ac ⁴

Respond:

$$\begin{gathered} \left(\mathrm{i}\right)5\mathrm{is}\mathrm{a}\mathrm{nonzero}\mathrm{constant}\mathrm{term}. \\ \mathrm{The}\mathrm{degree}\mathrm{of}\mathrm{a}\mathrm{nonzero}\mathrm{abiding}\mathrm{term}\mathrm{is}0\left(\mathrm{aught}\right). \\ \therefore \mathrm{Degree}\mathrm{of}5=0 \end{gathered}$$ $$\begin{gathered} \left(\mathrm{two}\right)-\frac{3}{\mathrm{two}}\mathrm{is}\mathrm{a}\mathrm{nonzero}\mathrm{abiding}\mathrm{term}. \\ \mathrm{The}\mathrm{degree}\mathrm{of}\mathrm{a}\mathrm{nonzero}\mathrm{constant}\mathrm{term}\mathrm{is}0\left(\mathrm{zero}\right). \\ \therefore \mathrm{Degree}\mathrm{of}-\frac{3}{\mathrm{two}}=0 \end{gathered}$$ $$\begin{gathered} \left(\mathrm{three}\right)4p-\sqrt{5}\mathrm{is}\mathrm{a}\mathrm{polynomial}\mathrm{in}\mathrm{variable}p. \\ \mathrm{The}\mathrm{highest}\mathrm{power}\mathrm{of}p\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{expression}\mathrm{is}1. \\ \therefore \mathrm{Degree}\mathrm{of}4p-\sqrt{5}=1 \end{gathered}$$ $$\begin{gathered} \left(\mathrm{iv}\right)3y+4y^2\mathrm{is}\mathrm{a}\mathrm{polynomial}\mathrm{in}\mathrm{variable}y. \\ \mathrm{The}\mathrm{highest}\mathrm{ability}\mathrm{of}y\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{expression}\mathrm{is}2. \\ \therefore \mathrm{Degree}\mathrm{of}3y+\mathrm{iv}{y}^{\mathrm{two}}=2 \end{gathered}$$ $$\begin{gathered} \left(\mathrm{v}\right)x^{\mathrm{ix}}-x^4+{\mathrm{ten}}^{12}+x-2\mathrm{is}\mathrm{a}\mathrm{polynomial}\mathrm{in}\mathrm{variable}x. \\ \mathrm{The}\mathrm{highest}\mathrm{power}\mathrm{of}x\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{expression}\mathrm{is}12. \\ \therefore \mathrm{Degree}\mathrm{of}{x}^9-x^4+x^{12}+\mathrm{ten}-\mathrm{ii}=12 \end{gathered}$$ $$\begin{gathered} \left(\mathrm{half\; dozen}\right)5g^{2}n\mathrm{is}\mathrm{a}\mathrm{polynomial}\mathrm{in}\mathrm{variable}s\mathrm{thousand}\mathrm{and}n. \\ \mathrm{Sum}\mathrm{of}\mathrm{the}\mathrm{power}s\mathrm{of}m\mathrm{and}n\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{expression}=\mathrm{two}+1=3 \\ \therefore \mathrm{Degree}\mathrm{of}5k^{\mathrm{ii}}\mathrm{due\; north}=\mathrm{iii} \end{gathered}$$ $$\begin{gathered} \left(\mathrm{vii}\right)2\mathrm{ten}y+4x^{\mathrm{iii}}\mathrm{is}\mathrm{a}\mathrm{polynomial}\mathrm{in}\mathrm{variables}x\mathrm{and}y. \\ \mathrm{The}\mathrm{maximum}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{powers}\mathrm{of}\mathrm{ten}\mathrm{and}y\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{expression}\mathrm{is}\mathrm{iii}. \\ \therefore \mathrm{Degree}\mathrm{of}\mathrm{ii}\mathrm{ten}y+\mathrm{four}{\mathrm{ten}}^{\mathrm{iii}}=3 \end{gathered}$$ $$\begin{gathered} \left(\mathrm{viii}\right)7p^{\mathrm{two}}{q}^{\mathrm{three}}t-11p^{4}t+2p^8\mathrm{is}\mathrm{a}\mathrm{polynomial}\mathrm{in}\mathrm{variable}p,q\mathrm{and}t. \\ \mathrm{The}\mathrm{maximum}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{powers}\mathrm{of}p,q\mathrm{and}t\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{expression}\mathrm{is}8. \\ \therefore \mathrm{Caste}\mathrm{of}7p^2{q}^{3}t-\mathrm{eleven}{p}^{4}t+\mathrm{two}{p}^8=\mathrm{viii} \end{gathered}$$ $$\begin{gathered} \left(ix\right)\mathrm{five}{\mathrm{10}}^{2}y{z}^{\mathrm{iii}}+x{y}^{\mathrm{iv}}{z}^2\mathrm{is}\mathrm{a}\mathrm{polynomial}\mathrm{in}\mathrm{variable}x,y\mathrm{and}z. \\ \mathrm{The}\mathrm{maximum}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{powers}\mathrm{of}x,y\mathrm{and}z\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{expression}\mathrm{is}7. \\ \therefore \mathrm{Degree}\mathrm{of}5x^{2}y{z}^3+\mathrm{10}{y}^4{z}^2=7 \end{gathered}$$ $$\begin{gathered} \left(\mathrm{x}\right)a{b}^{2}cd-3a^{2}bc{d}^{\mathrm{three}}+\mathrm{ii}a{c}^{\mathrm{iv}}\mathrm{is}\mathrm{a}\mathrm{polynomial}\mathrm{in}\mathrm{variables}a,b,c\mathrm{and}d. \\ \mathrm{The}\mathrm{maximum}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{powers}\mathrm{of}a,b,c\mathrm{and}d\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{expression}\mathrm{is}7. \\ \therefore \mathrm{Degree}\mathrm{of}a{b}^{2}cd-3a^{2}bc{d}^3+2a{c}^4=\mathrm{vii} \end{gathered}$$

Page No 53:

Question two:

Write the following polynomials in the standard class (descending type):
(i) 3x

$-$

x ⁴ + 5ten ³ + 2x ^two

$-$

5

(ii) 3x + five + 10 ³

(iii) 7 + 4t ² +

$\sqrt{5}$

t

(iv)

$\frac{\mathrm{eleven}}{\mathrm{viii}}{x}^{\mathrm{ii}}+2-\mathrm{seven}\mathrm{10}$

(5) 710 + ivx ² +

$\sqrt{\mathrm{three}}{\mathrm{ten}}^3-14$

Respond:

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{The}\mathrm{standard}\mathrm{form}\left(\mathrm{descending}\mathrm{type}\right)\mathrm{of}\mathrm{the}\mathrm{equation} \\ 3x-x^{\mathrm{iv}}+5{\mathrm{ten}}^3+2{\mathrm{ten}}^{\mathrm{two}}-\mathrm{v}\mathrm{will}\mathrm{exist} \\ -x^4+5x^{\mathrm{three}}+2x^2+3\mathrm{10}-5. \end{gathered}$$ $$\begin{gathered} \left(2\right)\mathrm{The}\mathrm{standard}\mathrm{form}\left(\mathrm{descending}\mathrm{blazon}\right)\mathrm{of}\mathrm{the}\mathrm{equation} \\ 3x+\mathrm{v}+x^3\mathrm{will}\mathrm{be} \\ {\mathrm{ten}}^{\mathrm{iii}}+3x+5. \end{gathered}$$ $$\begin{gathered} \left(\mathrm{iii}\right)\mathrm{The}\mathrm{standard}\mathrm{class}\left(\mathrm{descending}\mathrm{type}\right)\mathrm{of}\mathrm{the}\mathrm{equation} \\ 7+\mathrm{four}{t}^2+\sqrt{\mathrm{five}}t\mathrm{volition}\mathrm{exist} \\ 4t^{\mathrm{two}}+\sqrt{5}t+7. \end{gathered}$$ $$\begin{gathered} \left(\mathrm{iv}\right)\mathrm{The}\mathrm{standard}\mathrm{grade}\left(\mathrm{descending}\mathrm{blazon}\right)\mathrm{of}\mathrm{the}\mathrm{equation} \\ \frac{11}{8}{x}^2+2-7x\mathrm{volition}\mathrm{be} \\ \frac{11}{8}{x}^2-7x+2. \end{gathered}$$ $$\begin{gathered} \left(\mathrm{v}\right)\mathrm{The}\mathrm{standard}\mathrm{form}\left(\mathrm{descending}\mathrm{type}\right)\mathrm{of}\mathrm{the}\mathrm{equation} \\ 7x+4x^{\mathrm{two}}+\sqrt{3}{\mathrm{ten}}^3-\mathrm{xiv}\mathrm{will}\mathrm{exist} \\ \sqrt{3}{x}^{\mathrm{three}}+4{\mathrm{ten}}^2+7\mathrm{ten}-14. \end{gathered}$$

Page No 54:

Question 1:

Find the sum of the following polynomials and write the caste of the sum so obtained

(i) 210 ³

$-$

710 ² + 3x + 4 ; two10 ³

$-$

3ten ² + 4x + i
(ii) three10 ² + 5x

$-$

x ⁷ ;

$-$

3x ² + v10 + 8
(iii) x ⁴ + 5ten ³ + 7x ; 4x ³

$-$

3x ^two + 5
(four) y ² + iiy

$-$

five ; y ^three + 2y ² + 3y + four ; y ³ + 7y

$-$

2
(v) 5m ² + threem + 8 ; m ^three

$-$

sixthou ² + 4m ; m ³

$-$

thou ⁱⁱ

$-$

m + five

Reply:

$$\begin{gathered} \left(\mathrm{i}\right) \\ \mathrm{Required}\mathrm{polynomial}=\left(2x^3-\mathrm{vii}{x}^{\mathrm{two}}+\mathrm{three}\mathrm{ten}+\mathrm{iv}\right)+\left(2x^3-3x^2+4\mathrm{ten}+\mathrm{one}\right) \\ =\mathrm{two}{x}^3-7x^2+3\mathrm{10}+\mathrm{iv}+2x^3-\mathrm{three}{x}^{\mathrm{two}}+4\mathrm{ten}+1 \\ =2x^{\mathrm{iii}}+\mathrm{two}{x}^{\mathrm{iii}}-7x^{\mathrm{two}}-\mathrm{iii}{\mathrm{10}}^{\mathrm{two}}+\mathrm{three}x+4x+\mathrm{four}+\mathrm{ane} \\ =\mathrm{iv}{\mathrm{ten}}^3-\mathrm{ten}{x}^2+7x+\mathrm{five} \\ \mathrm{Here},\mathrm{the}\mathrm{highest}\mathrm{ability}\mathrm{is}\mathrm{iii}. \\ \therefore \mathrm{Degree}\mathrm{of}\mathrm{the}\mathrm{polynomial}=3 \end{gathered}$$ $$\begin{gathered} \left(2\right) \\ \mathrm{Required}\mathrm{polynomial}=\left(\mathrm{three}{x}^2+5x-{\mathrm{ten}}^{\mathrm{seven}}\right)+\left(-3x^{\mathrm{two}}+\mathrm{v}\mathrm{ten}+8\right) \\ =3x^2+5\mathrm{ten}-x^7-3x^2+5x+8 \\ =-{\mathrm{ten}}^7+3x^{\mathrm{ii}}-3x^2+\mathrm{five}x+5x+8 \\ =-x^{\mathrm{seven}}+10x+8 \\ \mathrm{Here},\mathrm{the}\mathrm{highest}\mathrm{power}\mathrm{is}7. \\ \therefore \mathrm{Degree}\mathrm{of}\mathrm{the}\mathrm{polynomial}=\mathrm{seven} \end{gathered}$$ $$\begin{gathered} \left(\mathrm{iii}\right) \\ \mathrm{Required}\mathrm{polynomial}=\left(x^4+5{\mathrm{ten}}^3+\mathrm{vii}\mathrm{10}\right)+\left(\mathrm{iv}{x}^3-\mathrm{three}{x}^2+\mathrm{v}\right) \\ =x^4+5x^3+\mathrm{vii}\mathrm{ten}+\mathrm{four}{x}^{\mathrm{iii}}-3{\mathrm{10}}^2+\mathrm{five} \\ =x^{\mathrm{iv}}+5{\mathrm{10}}^3+\mathrm{iv}{x}^3-\mathrm{three}{\mathrm{ten}}^2+7x+5 \\ =x^4+\mathrm{ix}{x}^3-3{\mathrm{ten}}^{\mathrm{ii}}+7x+5 \\ \mathrm{Here},\mathrm{the}\mathrm{highest}\mathrm{power}\mathrm{is}\mathrm{four}. \\ \therefore \mathrm{Degree}\mathrm{of}\mathrm{the}\mathrm{polynomial}=\mathrm{four} \end{gathered}$$ $$\begin{gathered} \left(4\right) \\ \mathrm{Required}\mathrm{polynomial}=\left(y^{\mathrm{two}}+\mathrm{two}y-5\right)+\left(y^3+2y^{\mathrm{two}}+3y+4\right)+\left(y^{\mathrm{iii}}+\mathrm{vii}y-2\right) \\ =y^{\mathrm{two}}+2y-5+y^3+\mathrm{two}{y}^{\mathrm{two}}+\mathrm{three}y+4+y^{\mathrm{iii}}+7y-2 \\ =y^{\mathrm{three}}+y^3+y^2+2y^{\mathrm{ii}}+2y+\mathrm{three}y+7y-5+\mathrm{iv}-2 \\ =\mathrm{ii}{y}^{\mathrm{three}}+3y^2+12y-\mathrm{three} \\ \mathrm{Hither},\mathrm{the}\mathrm{highest}\mathrm{power}\mathrm{is}3. \\ \therefore \mathrm{Degree}\mathrm{of}\mathrm{the}\mathrm{polynomial}=3 \end{gathered}$$ $$\begin{gathered} \left(\mathrm{5}\right) \\ \mathrm{Required}\mathrm{polynomial}=\left(5m^{\mathrm{two}}+\mathrm{iii}m+8\right)+\left({\mathrm{thou}}^{\mathrm{iii}}-\mathrm{six}{g}^2+\mathrm{four}\mathrm{grand}\right)+\left(k^3-{\mathrm{thousand}}^{\mathrm{ii}}-\mathrm{grand}+5\right) \\ =5m^{\mathrm{two}}+3m+\mathrm{eight}+m^3-6m^2+\mathrm{iv}\mathrm{thou}+m^{\mathrm{three}}-{\mathrm{yard}}^2-m+5 \\ ={\mathrm{1000}}^3+m^{\mathrm{iii}}+5m^{\mathrm{ii}}-6m^2-m^{\mathrm{two}}+3\mathrm{1000}+\mathrm{four}m-m+8+5 \\ =2{\mathrm{one\; thousand}}^3-2m^{\mathrm{ii}}+\mathrm{half\; dozen}\mathrm{grand}+\mathrm{thirteen} \\ \mathrm{Here},\mathrm{the}\mathrm{highest}\mathrm{power}\mathrm{is}3. \\ \therefore \mathrm{Caste}\mathrm{of}\mathrm{the}\mathrm{polynomial}=3 \end{gathered}$$

Page No 54:

Question two:

Subtract the 2nd polynomial from the first and write the caste of the polynomial so obtained.

(i) x ⁴ + x ² + x

$-$

ane ; 10 ^iv

$-$

x ³

$-$

x ²

$-$

i
(ii) n ^three

$-$

5northward ² + six ; north ⁱⁱ

$-$

3north + 8
(three) twoa + iiia ²

$-$

seven ; 3a ⁱⁱ

$-$

12 + 2a

Answer:

$$\begin{gathered} \left(\mathrm{i}\right) \\ \mathrm{Required}\mathrm{polynomial}=\left({\mathrm{ten}}^4+x^{\mathrm{two}}+x-\mathrm{ane}\right)-\left(x^4-x^{\mathrm{iii}}-x^{\mathrm{two}}+\mathrm{one}\right) \\ ={\mathrm{ten}}^{\mathrm{iv}}+{\mathrm{ten}}^{\mathrm{two}}+x-1-x^{\mathrm{iv}}+{\mathrm{ten}}^{\mathrm{iii}}+x^2-1 \\ ={\mathrm{ten}}^4-x^{\mathrm{four}}+{\mathrm{10}}^3+x^{\mathrm{two}}+x^2+\mathrm{ten}-1-\mathrm{ane} \\ ={\mathrm{ten}}^3+2x^2+x-2 \\ \mathrm{Here},\mathrm{the}\mathrm{highest}\mathrm{power}\mathrm{is}3. \\ \therefore \mathrm{Degree}\mathrm{of}\mathrm{the}\mathrm{polynomial}=3 \end{gathered}$$ $$\begin{gathered} \left(\mathrm{ii}\right) \\ \mathrm{Required}\mathrm{polynomial}=\left(n^3-5{\mathrm{north}}^2+\mathrm{half\; dozen}\right)-\left(n^3-\mathrm{iii}\mathrm{north}+8\right) \\ =n^3-\mathrm{five}{n}^2+6-n^3+3n-\mathrm{viii} \\ ={\mathrm{northward}}^{\mathrm{iii}}-n^{\mathrm{three}}-\mathrm{five}{n}^{\mathrm{ii}}+3n+\mathrm{six}-\mathrm{eight} \\ =-5n^2+3n-2 \\ \mathrm{Here},\mathrm{the}\mathrm{highest}\mathrm{power}\mathrm{is}\mathrm{ii}. \\ \therefore \mathrm{Degree}\mathrm{of}\mathrm{the}\mathrm{polynomial}=\mathrm{ii} \end{gathered}$$ $$\begin{gathered} \left(3\right) \\ \mathrm{Required}\mathrm{polynomial}=\left(\mathrm{two}a+3a^2-\mathrm{vii}\right)-\left(3a^2-12+2a\right) \\ =\mathrm{two}a+\mathrm{three}{a}^2-\mathrm{vii}-3a^2+12-2a \\ =\mathrm{three}{a}^{\mathrm{two}}-3a^2+2a-\mathrm{two}a-7+12 \\ =5 \\ \mathrm{Here},\mathrm{the}\mathrm{polynomial}\mathrm{is}\mathrm{a}\mathrm{nonzero}\mathrm{constant}. \\ \mathrm{W}e\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{degree}\mathrm{of}\mathrm{a}\mathrm{nonzero}\mathrm{constant}\mathrm{is}0. \\ \therefore \mathrm{Degree}\mathrm{of}\mathrm{the}\mathrm{polynomial}=0 \end{gathered}$$

Page No 54:

Question 3:

Simplify:

(i) (iiix ²

$-$

twox + 1) + (x ^two + 5ten

$-$

3) + (410 ² + viii)
(ii) (2y ^three + 3y

$-$

7)

$-$

(8y

$-$

6) + (4y ³

$-$

iiy + 1)
(iii) 5m ³

$-$

m + half-dozenyard ²

$-$

(3m ²

$-$

2 + m)

Answer:

$$\begin{gathered} \left(\mathrm{i}\right)\left(3x^2-2x+\mathrm{i}\right)+\left({\mathrm{ten}}^{\mathrm{ii}}+5\mathrm{10}-\mathrm{iii}\right)+\left(4x^2+8\right) \\ =3{\mathrm{ten}}^{\mathrm{two}}-\mathrm{two}x+\mathrm{ane}+x^{\mathrm{two}}+5x-3+\mathrm{four}{x}^2+8 \\ =\mathrm{iii}{\mathrm{10}}^2+x^2+\mathrm{four}{\mathrm{10}}^{\mathrm{two}}-\mathrm{two}x+5\mathrm{ten}+\mathrm{i}-3+8 \\ =\mathrm{viii}{x}^{\mathrm{two}}+3x+\mathrm{vi} \end{gathered}$$ $$\begin{gathered} \left(\mathrm{ii}\right)\left(2y^3+\mathrm{three}y-7\right)-\left(8y-6\right)+\left(\mathrm{four}{y}^3-\mathrm{ii}y+1\right) \\ =2y^3+\mathrm{iii}y-\mathrm{vii}-8y+\mathrm{vi}+\mathrm{four}{y}^3-\mathrm{ii}y+1 \\ =2y^3+4y^{\mathrm{iii}}+\mathrm{three}y-8y-\mathrm{two}y-7+6+1 \\ =\mathrm{six}{y}^3-7y \end{gathered}$$ $$\begin{gathered} \left(\mathrm{iii}\right)5m^{\mathrm{iii}}-m+\mathrm{half\; dozen}{m}^{\mathrm{ii}}-\left(\mathrm{three}{m}^2-2+\mathrm{thou}\right) \\ =5m^3-\mathrm{thousand}+\mathrm{half-dozen}{\mathrm{chiliad}}^2-3m^2+2-\mathrm{1000} \\ =\mathrm{v}{m}^3+6g^2-3m^{\mathrm{ii}}-\mathrm{thou}-m+2 \\ =\mathrm{five}{\mathrm{thousand}}^{\mathrm{three}}+3k^2-\mathrm{ii}\mathrm{yard}+\mathrm{ii} \end{gathered}$$

Page No 54:

Question iv:

Which polynomial should exist added to twox ⁴ $-$ 310 ⁱⁱ + 5x + eight to become twox ² $-$ 5x + four?

Answer:

$$\begin{gathered} \mathrm{Allow}\mathrm{the}\mathrm{polynomial}\mathrm{to}\mathrm{be}\mathrm{added}\mathrm{be}p\left(x\right). \\ \mathrm{Then}\left(2x^4-\mathrm{iii}{x}^2+\mathrm{five}\mathrm{10}+8\right)+p\left(\mathrm{ten}\right)=\left(2x^2-5\mathrm{10}+4\right) \\ \Rightarrow p\left(x\right)=\left(2x^2-5x+4\right)-\left(2{\mathrm{ten}}^4-3{\mathrm{ten}}^{\mathrm{ii}}+5x+\mathrm{eight}\right) \\ =2{\mathrm{10}}^2-\mathrm{v}x+4-\mathrm{ii}{\mathrm{ten}}^{\mathrm{iv}}+3{\mathrm{ten}}^2-5\mathrm{10}-8 \\ =-2x^4+2x^{\mathrm{two}}+3{\mathrm{ten}}^2-\mathrm{five}\mathrm{10}-\mathrm{v}x+\mathrm{four}-\mathrm{viii} \\ =-2x^{\mathrm{four}}+\mathrm{v}{x}^2-10x-\mathrm{four} \\ \therefore -2x^4+5{\mathrm{10}}^2-\mathrm{x}x-\mathrm{four}\mathrm{should}\mathrm{be}\mathrm{added}\mathrm{to}2x^{\mathrm{iv}}-3{\mathrm{10}}^{\mathrm{two}}+\mathrm{v}\mathrm{10}+\mathrm{viii}\mathrm{to}\mathrm{get}\mathrm{two}{x}^2-5x+4. \end{gathered}$$

Folio No 54:

Question 5:

Which polynomial should exist subtracted  from y ³ + 2y ^two + 5y − 1 to get iiy ² + 12?

Answer:

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{polynomial}\mathrm{to}\mathrm{exist}\mathrm{subtracted}\mathrm{be}p\left(x\right). \\ \mathrm{So}\left(y^3+2y^{\mathrm{two}}+\mathrm{five}y-1\right)-p\left(x\right)=\left(2y^{\mathrm{two}}+12\right) \\ \Rightarrow p\left(x\right)=\left(y^{\mathrm{iii}}+\mathrm{two}{y}^{\mathrm{two}}+5y-1\right)-\left(2y^{\mathrm{ii}}+12\right) \\ =y^3+2y^2+5y-1-\mathrm{two}{y}^2-12 \\ =y^3+\mathrm{two}{y}^2-\mathrm{two}{y}^2+\mathrm{v}y-1-12 \\ =y^3+\mathrm{five}y-13 \\ \therefore y^3+5y-13\mathrm{should}\mathrm{be}\mathrm{subtracted}\mathrm{from}{y}^{\mathrm{three}}+\mathrm{two}{y}^2+5y-\mathrm{one}\mathrm{to}\mathrm{get}2y^2+12. \end{gathered}$$

Page No 54:

Question 6:

From the sum of z ³ + 3z ^two + fivez + 8  and fourz ³ + twoz ⁱⁱ $-$ 7z $-$ ii subtract iix ^three $-$ 3z ² + z $-$ four.

Answer:

$$\begin{gathered} \mathrm{Required}\mathrm{polynomial}=\left(z^{\mathrm{iii}}+\mathrm{three}{z}^{\mathrm{two}}+5z+8\right)+\left(4z^3+2z^2-7z-2\right)-\left(\mathrm{ii}{z}^3-3z^2+z-4\right) \\ =z^3+\mathrm{iii}{z}^{\mathrm{two}}+\mathrm{v}z+8+4z^{\mathrm{iii}}+2z^{\mathrm{ii}}-7z-2-\mathrm{two}{z}^3+3z^{\mathrm{two}}-z+4 \\ =z^3+4z^{\mathrm{iii}}-\mathrm{ii}{z}^3+\mathrm{three}{z}^2+2z^2+3z^{\mathrm{ii}}+\mathrm{v}z-7z-z+8-\mathrm{ii}+4 \\ =\mathrm{three}{z}^3+8z^2-\mathrm{iii}z+10 \end{gathered}$$

Page No 56:

Question 1:

Find the product of the following polynomials and state the degree of their product.

(i) x ² + iii10 + 1; 2x

$-$

3
(ii) 3x ^two + 5x ; 10 ² + ii10 + 1
(iii) x ⁱⁱⁱ + 4x + 2 ; x ² + x + v
(four) ten ⁱⁱⁱ

$-$

1 ; x ²

$-$

x + 4
(v) 2y ^two + three ; 3y ^three + 1

Answer:

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{We}\mathrm{have}\left(x^2+3\mathrm{ten}+1\right)\left(2x-3\right) \\ =x^2\left(2\mathrm{10}-3\right)+3x\left(\mathrm{ii}\mathrm{10}-\mathrm{three}\right)+1\left(2x-3\right) \\ =2x^3-3x^{\mathrm{ii}}+6x^{\mathrm{ii}}-9x+2\mathrm{10}-\mathrm{iii} \\ =2x^3+3x^2-\mathrm{vii}\mathrm{10}-3 \\ \therefore \mathrm{Degree}=3 \end{gathered}$$ $$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{We}\mathrm{have}\left(\mathrm{three}{\mathrm{ten}}^2+5x\right)\left({\mathrm{10}}^2+\mathrm{ii}x+\mathrm{i}\right) \\ =3x^2\left({\mathrm{10}}^2+2x+1\right)+5\mathrm{10}\left(x^2+\mathrm{two}\mathrm{10}+\mathrm{i}\right) \\ =3x^{\mathrm{four}}+\mathrm{six}{x}^{\mathrm{three}}+\mathrm{three}{\mathrm{ten}}^2+5x^3+10{\mathrm{ten}}^2+5x \\ =3{\mathrm{10}}^{\mathrm{iv}}+11x^3+13{\mathrm{10}}^2+5x \\ \therefore \mathrm{Degree}=4 \end{gathered}$$ $$\begin{gathered} \left(\mathrm{iii}\right)\mathrm{Nosotros}\mathrm{have}\left(x^{\mathrm{iii}}+4\mathrm{ten}+2\right)\left(x^{\mathrm{two}}+\mathrm{10}+5\right) \\ =x^{\mathrm{iii}}\left(x^{\mathrm{ii}}+\mathrm{10}+5\right)+4x\left(x^2+x+5\right)+\mathrm{ii}\left({\mathrm{10}}^2+x+\mathrm{v}\right) \\ =x^5+x^{\mathrm{iv}}+\mathrm{v}{x}^{\mathrm{three}}+4x^{\mathrm{three}}+\mathrm{iv}{x}^2+20x+2{\mathrm{ten}}^{\mathrm{two}}+\mathrm{two}x+\mathrm{x} \\ =x^5+x^4+9{\mathrm{10}}^{\mathrm{three}}+\mathrm{six}{x}^{\mathrm{ii}}+22x+10 \\ \therefore \mathrm{Degree}=\mathrm{five} \end{gathered}$$ $$\begin{gathered} \left(\mathrm{iv}\right)\mathrm{We}\mathrm{have}\left({\mathrm{10}}^3-1\right)\left(x^2-x+4\right) \\ =x^{\mathrm{three}}\left({\mathrm{ten}}^2-x+4\right)-1\left({\mathrm{10}}^2-x+4\right) \\ =x^{\mathrm{five}}-{\mathrm{ten}}^4+4x^{\mathrm{iii}}-x^2+x-4 \\ \therefore \mathrm{Degree}=5 \end{gathered}$$ $$\begin{gathered} \left(\mathrm{v}\right)\mathrm{We}\mathrm{have}\left(2y^2+3\right)\left(\mathrm{three}{y}^3+\mathrm{ane}\right) \\ =\mathrm{two}{y}^2\left(3y^3+\mathrm{ane}\right)+3\left(\mathrm{three}{y}^3+\mathrm{i}\right) \\ =\mathrm{half-dozen}{y}^5+\mathrm{ii}{y}^{\mathrm{two}}+\mathrm{nine}{y}^3+3 \\ =\mathrm{half-dozen}{y}^5+\mathrm{ix}{y}^{\mathrm{three}}+2y^2+3 \\ \therefore \mathrm{Degree}=5 \end{gathered}$$

Folio No 56:

Question ii.1:

In each of the following case divide the start polynomial by the second polynomial and express every bit Divident = Divisor ✕ quotient + Remainder.

(i) x ³

$-$

v10 ² + 4ten + 8 ; 10 + 2

Answer:

Here, caliber = x ²

$-$

710 + xviii
and remainder =

$-$

28

$\therefore x^3-\mathrm{v}{x}^{\mathrm{two}}+4x+\mathrm{eight}=\left(x+2\right)\left(x^2-7\mathrm{ten}+18\right)-28$

Folio No 56:

Question 2.2:

y ³ $-$ sixy ² + 6y + 1 ; y $-$ 1

Answer:

Here, caliber = y ⁱⁱ

$-$

5y + ane
and residue = 2

$\therefore y^3-\mathrm{six}{y}^2+6y+1=\left(y-\mathrm{one}\right)\left(y^{\mathrm{ii}}-5y+\mathrm{i}\right)+2$

Page No 56:

Question two.3:

y ^three $-$ 64 ; y $-$ 4

Answer:

Hither, caliber = y ² + ivy + 16
and residual = 0

$\therefore y^3-64=\left(y-\mathrm{four}\right)\left(y^2+\mathrm{iv}y+16\right)+0$

Folio No 56:

Question 2.4:

6x ^three + fiveten ² $-$ 21x + ten, 3x $-$ 2

Answer:

Here, caliber = 2x ⁱⁱ + 3ten

$-$

5
and remainder = 0

$\therefore \mathrm{vi}{\mathrm{ten}}^3+5x^2-21x+10=\left(3x-2\right)\left(\mathrm{ii}{\mathrm{ten}}^2+\mathrm{three}\mathrm{10}-\mathrm{five}\right)+0$

Page No 56:

Question ii.5:

 3ten ^five $-$ fourx ^iv + 3x ⁱⁱⁱ + 210 ; 10 ⁱⁱ $-$ three

Answer:

Here, quotient = 3x ³

$-$

4x ² + 12ten

$-$

12
and residuum = 3810

$-$

36

$\therefore 3{\mathrm{ten}}^{\mathrm{five}}-4x^{\mathrm{iv}}+3{\mathrm{ten}}^{\mathrm{iii}}+2\mathrm{ten}=\left({\mathrm{ten}}^{\mathrm{two}}-3\right)\left(3{\mathrm{ten}}^{\mathrm{three}}-\mathrm{four}{\mathrm{10}}^{\mathrm{ii}}+12\mathrm{ten}-12\right)+38x-36$

Page No 59:

Question 1:

Express the following polynomials in the coefficient form:

(i) 2ten ^two + fivex + 12
(ii) y ⁴

$-$

3y ² + 2y

$-$

7
(iii) ten ⁵ + iiix ⁱⁱ
(iv) y ^iv

$-$

3
(v) ninex

Answer:

(i) The degree of the given polynomial is 2.
∴ Number of terms in the alphabetize course of the polynomial = ii + 1 = 3
The polynomial

$\mathrm{two}{x}^{\mathrm{two}}+5\mathrm{ten}+12$

tin can be written in the alphabetize form as

$2x^2+5x+12$

.
∴ The coefficient course of the given polynomial is (2, 5, 12).

(ii) The degree of the given polynomial is 4.
∴ Number of terms in the index class of the polynomial = iv + 1 = 5
The polynomial

$y^4-3y^2+2y-\mathrm{seven}$

can exist written in the index form every bit

$y^{\mathrm{iv}}+0y^{\mathrm{three}}-3y^{\mathrm{ii}}+2y-\mathrm{vii}$

.
∴ The coefficient class of the given polynomial is (one, 0,

$-$

3, 2,

$-$

7).

(iii) The caste of the given polynomial is 5.
∴ Number of terms in the index form of the polynomial = 5 + 1 = 6
The polynomial

$x^{\mathrm{v}}+3x^2$

can be written in the index form as

$x^{\mathrm{five}}+0x^{\mathrm{iv}}+0{\mathrm{ten}}^3+3x^2+0\mathrm{10}+0$

.
∴ The coefficient course of the given polynomial is (one, 0, 0,  3, 0, 0).

(iv) The degree of the given polynomial is four.
∴ Number of terms in the index form of the polynomial = iv + one = 5
The polynomial

$y^4-3$

can be written in the index form as

$y^4+0y^3+0y^2+0y-3$

.
∴ The coefficient course of the given polynomial is (i, 0, 0,  0,

$-$

3).

(v) The degree of the given polynomial is one.
∴ Number of terms in the index form of the polynomial = 1 + i = 2
The polynomial

$9x$

tin be written in the index course as

$9\mathrm{ten}+0$

∴ The coefficient course of the given polynomial is (1, 0).

Folio No 59:

Question 2:

Express the following polynomials in the index form taking 'x' equally a variable.

(i) (iii, 2, 7)
(ii) (2, 0, 0,

$-$

4)
(iii) (one, 0,

$-$

3, 1, five)
(4) (

$-$

2, 3,

$-$

5, six)
(five) ( ane, 0, 0, 0, 0, 0, 64)

Answer:

(i) The polynomial (3, 2, vii) contains 3 coefficients.
i.e., degree of the polynomial = 3

$-$

one = 2
∴ The index class of the given polynomial is 3x ² + iiten + vii.

(ii) The polynomial (2, 0, 0,

$-$

4) contains 4 coefficients.
i.e., degree of the polynomial = 4

$-$

one = 3
∴ The index form of the given polynomial is 210 ³ + 0 x ² + 0 10

$-$

four.

(iii) The polynomial (i, 0,

$-$

3, 1, five) contains five coefficients.
i.east., degree of the polynomial = 5

$-$

1 = four
∴ The index form of the given polynomial is x ⁴ + 0 x ³

$-$

3 x ^two + x + five.

(four) The polynomial (

$-$

ii, 3,

$-$

5, six) contains 4 coefficients.
i.e., degree of the polynomial = 4

$-$

ane = 3
∴ The index form of the given polynomial is

$-$

2x ³ + 3ten ⁱⁱ

$-$

5 x + six.

(v) The polynomial (1, 0, 0, 0, 0, 0, 64) contains seven coefficients.
i.due east., degree of the polynomial = 7

$-$

1 = vi
∴ The index form of the given polynomial is x ^vi + 0 10 ^v + 0 10 ⁴ + 0 x ³ + 0 ten ^two + 0 ten + 64.

Page No 59:

Question 3.one:

Use constructed division method for performing the following divisions. Write the result in the form
Divident = Divisor

$\times$

Quotient + Rest.

(i) (x ³

$-$

ivx ⁱⁱ

$-$

2x + 1)

$÷$

(10

$-$

3)

Answer:

$$\begin{gathered} \mathrm{Dividend}={\mathrm{ten}}^3-\mathrm{iv}{x}^2-2\mathrm{ten}+\mathrm{i} \\ \mathrm{Index}\mathrm{class}=x^3-4x^2-\mathrm{two}x+\mathrm{i} \\ \mathrm{Coefficient}\mathrm{form}=\left(\mathrm{i},-\mathrm{four},-\mathrm{ii},1\right) \\ \mathrm{Comparison}\mathrm{the}\mathrm{divisor}x-3\mathrm{with}x-a,\mathrm{we}\mathrm{get}a=\mathrm{iii} \end{gathered}$$

$$\begin{gathered} \mathrm{At\; present},\mathrm{caliber}\mathrm{in}\mathrm{coefficient}\mathrm{form}\mathrm{is}\left(\mathrm{ane},-1,-\mathrm{v}\right) \\ \therefore \mathrm{Caliber}={\mathrm{10}}^{\mathrm{two}}-\mathrm{ten}-5\mathrm{in}\mathrm{the}\mathrm{variable}x \\ \mathrm{Remainder}=-\mathrm{xiv} \\ \therefore x^3-4x^{\mathrm{two}}-2x+\mathrm{ane}=\left(\mathrm{10}-3\right)\left(x^{\mathrm{ii}}-x-\mathrm{five}\right)-14 \end{gathered}$$

Folio No 59:

Question 3.2:

(two10 ³ $-$ iiix ^two + 4x + ii) $÷$ (ten $-$ 1)

Answer:

$$\begin{gathered} \mathrm{Dividend}=\mathrm{two}{\mathrm{ten}}^3-3x^2+4\mathrm{10}+2 \\ \mathrm{Index}\mathrm{grade}=2x^3-3x^2+4\mathrm{10}+2 \\ \mathrm{Coefficient}\mathrm{form}=\left(2,-\mathrm{three},\mathrm{four},\mathrm{ii}\right) \\ \mathrm{Comparing}\mathrm{the}\mathrm{divisor}\mathrm{ten}-1\mathrm{with}x-a,\mathrm{we}\mathrm{become}a=\mathrm{one} \end{gathered}$$

$$\begin{gathered} \mathrm{At\; present},\mathrm{quotient}\mathrm{in}\mathrm{coefficient}\mathrm{class}=\left(\mathrm{two},-\mathrm{ane},3\right) \\ \therefore \mathrm{Caliber}=\mathrm{ii}{x}^2-x+3\mathrm{in}\mathrm{the}\mathrm{variable}x \\ \mathrm{Remainder}=5 \\ \therefore 2x^3-\mathrm{iii}{\mathrm{ten}}^2+4x+\mathrm{two}=\left(\mathrm{ten}-\mathrm{ane}\right)\left(2x^{\mathrm{ii}}-x+\mathrm{iii}\right)+5 \end{gathered}$$

Folio No 59:

Question three.3:

(y ³ + 343) $÷$ (y + 7)

Reply:

$$\begin{gathered} \mathrm{Dividend}=y^3+343 \\ \mathrm{Index}\mathrm{form}=y^3+0y^2+0y+343 \\ \mathrm{Coefficient}\mathrm{form}=\left(1,0,0,343\right) \\ \mathrm{Comparing}\mathrm{the}\mathrm{divisor}y+\mathrm{vii}\mathrm{with}y-a,\mathrm{nosotros}\mathrm{get}a=-\mathrm{seven} \end{gathered}$$

$$\begin{gathered} \mathrm{At\; present},\mathrm{caliber}\mathrm{in}\mathrm{coefficient}\mathrm{form}=\left(1,-7,49\right) \\ \therefore \mathrm{Quotient}=y^2-7y+49\mathrm{in}\mathrm{the}\mathrm{variable}y \\ \mathrm{Rest}=0 \\ \therefore y^3+343=\left(y+7\right)\left(y^{\mathrm{two}}-7y+49\right)+0 \end{gathered}$$

Page No 59:

Question 3.4:

(x ⁵ + 10 ³ + x ^two $-$ two10 + 4) $÷$ (x + 3)

Reply:

$$\begin{gathered} \mathrm{Dividend}={\mathrm{10}}^5+x^{\mathrm{three}}+x^{\mathrm{two}}-\mathrm{two}\mathrm{10}+4 \\ \mathrm{Index}\mathrm{form}=x^5+0x^4+x^3+x^2-2x+4 \\ \mathrm{Coefficient}\mathrm{form}=\left(1,0,\mathrm{ane},\mathrm{i},-\mathrm{ii},\mathrm{four}\right) \\ \mathrm{Comparing}\mathrm{the}\mathrm{divisor}\mathrm{ten}+3\mathrm{with}x-a,\mathrm{we}\mathrm{get}a=-3 \end{gathered}$$

$$\begin{gathered} \mathrm{Now},\mathrm{quotient}\mathrm{in}\mathrm{coefficient}\mathrm{form}=\left(1,-3,10,-29,85\right) \\ \therefore \mathrm{Quotient}=x^4-\mathrm{iii}{x}^3+10x^2-29\mathrm{10}+85\mathrm{in}\mathrm{the}\mathrm{variable}x \\ \mathrm{Remainder}=-251 \\ \therefore x^{\mathrm{v}}+x^3+x^2-2x+\mathrm{iv}=\left(\mathrm{10}+3\right)\left(x^{\mathrm{iv}}-\mathrm{three}{\mathrm{ten}}^{\mathrm{three}}+\mathrm{ten}{x}^{\mathrm{ii}}-29x+85\right)-251 \end{gathered}$$

Page No 59:

Question 3.5:

(x ^three + 210 ² + x + two) $÷$ (10 $-$ 1)

Answer:

$$\begin{gathered} \mathrm{Dividend}=x^{\mathrm{three}}+2x^{\mathrm{ii}}+x+\mathrm{two} \\ \mathrm{Index}\mathrm{form}=x^3+2x^2+x+\mathrm{two} \\ \mathrm{Coefficient}\mathrm{form}=\left(1,2,1,2\right) \\ \mathrm{Comparing}\mathrm{the}\mathrm{divisor}x-1\mathrm{with}\mathrm{10}-a,\mathrm{nosotros}\mathrm{get}a=\mathrm{i} \end{gathered}$$

$$\begin{gathered} \mathrm{Now},\mathrm{quotient}\mathrm{in}\mathrm{coefficient}\mathrm{grade}=\left(\mathrm{ane},3,4\right) \\ \therefore \mathrm{Caliber}={\mathrm{10}}^{\mathrm{ii}}+3\mathrm{10}+4\mathrm{in}\mathrm{the}\mathrm{variable}x \\ \mathrm{Remainder}=6 \\ \therefore x^{\mathrm{iii}}+\mathrm{two}{\mathrm{10}}^2+\mathrm{ten}+\mathrm{ii}=\left(x-\mathrm{i}\right)\left(x^{\mathrm{ii}}+3x+4\right)+6 \end{gathered}$$

Page No 59:

Question three.6:

(y ² $-$ xiy + thirty)$÷$ (y $-$ 5)

Respond:

$$\begin{gathered} \mathrm{Dividend}=y^2-\mathrm{xi}y+30 \\ \mathrm{Index}\mathrm{grade}=y^2-\mathrm{eleven}y+30 \\ \mathrm{Coefficient}\mathrm{form}=\left(1,-11,30\right) \\ \mathrm{Comparing}\mathrm{the}\mathrm{divisor}y-\mathrm{five}\mathrm{with}y-a,\mathrm{we}\mathrm{get}a=5 \end{gathered}$$

$$\begin{gathered} \mathrm{Now},\mathrm{quotient}\mathrm{in}\mathrm{coefficient}\mathrm{form}=\left(1,-6\right) \\ \therefore \mathrm{Caliber}=y-\mathrm{six}\mathrm{in}\mathrm{the}\mathrm{variable}y \\ \mathrm{Remainder}=0 \\ \therefore y^{\mathrm{two}}-\mathrm{eleven}y+\mathrm{thirty}=\left(y-5\right)\left(y-\mathrm{six}\right)+0 \end{gathered}$$

Page No 59:

Question iii.7:

(x ⁱⁱⁱ $-$ 3x ² $-$ 12x + iv) $÷$ (10 $-$ 2)

Reply:

$$\begin{gathered} \mathrm{Dividend}=x^{\mathrm{iii}}-3x^{\mathrm{ii}}-12x+4 \\ \mathrm{Alphabetize}\mathrm{form}=x^{\mathrm{three}}-3x^2-12x+4 \\ \mathrm{Coefficient}\mathrm{course}=\left(\mathrm{ane},-3,-12,4\right) \\ \mathrm{Comparing}\mathrm{the}\mathrm{divisor}\mathrm{ten}-\mathrm{ii}\mathrm{with}\mathrm{10}-a,\mathrm{we}\mathrm{go}a=2 \end{gathered}$$

$$\begin{gathered} \mathrm{At\; present},\mathrm{quotient}\mathrm{in}\mathrm{coefficient}\mathrm{form}=\left(\mathrm{one},-1,-\mathrm{fourteen}\right) \\ \therefore \mathrm{Quotient}=x^2-x-\mathrm{xiv}\mathrm{in}\mathrm{the}\mathrm{variable}\mathrm{ten} \\ \mathrm{Remainder}=-24 \\ \therefore x^{\mathrm{three}}-3x^2-12x+4=\left(\mathrm{10}-2\right)\left(x^2-x-\mathrm{fourteen}\right)-24 \end{gathered}$$

Page No 59:

Question 3.8:

(2x ^iv + iiiten ² + v) $÷$ (10 + ii)

Respond:

$$\begin{gathered} \mathrm{Dividend}=2x^{\mathrm{four}}+\mathrm{three}{x}^2+5 \\ \mathrm{Alphabetize}\mathrm{form}=2x^4+0{\mathrm{ten}}^3+3x^2+0\mathrm{10}+5 \\ \mathrm{Coefficient}\mathrm{form}=\left(\mathrm{two},0,\mathrm{iii},0,\mathrm{five}\right) \\ \mathrm{Comparing}\mathrm{the}\mathrm{divisor}x+\mathrm{two}\mathrm{with}\mathrm{10}-a,\mathrm{nosotros}\mathrm{get}a=-2 \end{gathered}$$

$$\begin{gathered} \mathrm{At\; present},\mathrm{the}\mathrm{quotient}\mathrm{in}\mathrm{coefficient}\mathrm{grade}\mathrm{is}\left(2,-4,11,-22\right). \\ \therefore \mathrm{Caliber}=2x^{\mathrm{three}}-4{\mathrm{10}}^2+11x-22\mathrm{in}\mathrm{the}\mathrm{variable}x \\ \mathrm{Remainder}=49 \\ \therefore 2x^4+3{\mathrm{10}}^2+5=\left(x+2\right)\left(2x^3-4{\mathrm{ten}}^2+\mathrm{eleven}\mathrm{10}-22\right)+49 \end{gathered}$$

Page No lx:

Question ane:

Find the value of the polynomial x ² + 2x + five when,

(i) 10 = 0
(ii) x = 3
(iii) ten =

$-$

ane
(iv) ten =

$-$

3
(five) x = a

Answer:

Let p(x) = x ² + 210 + five

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{To}\mathrm{finding}\mathrm{the}\mathrm{value}\mathrm{of}p\left(\mathrm{10}\right)\mathrm{when}\mathrm{ten}=0,\mathrm{put}x=0\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{polynomial}. \\ p\left(\mathrm{10}\right)=x^{\mathrm{two}}+\mathrm{two}x+\mathrm{five} \\ \Rightarrow p\left(0\right)=0^{\mathrm{ii}}+\mathrm{ii}\times 0+5 \\ =5 \\ \Rightarrow p\left(0\right)=5 \\ \therefore \mathrm{The}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{polynomial}\mathrm{is}5\mathrm{when}x=0 \end{gathered}$$ $$\begin{gathered} \left(\mathrm{two}\right)\mathrm{To}\mathrm{finding}\mathrm{the}\mathrm{value}\mathrm{of}p\left(x\right)\mathrm{when}\mathrm{ten}=3,\mathrm{put}x=3\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{polynomial}. \\ p\left(x\right)=x^2+\mathrm{ii}x+5 \\ \Rightarrow p\left(\mathrm{three}\right)=3^2+2\times 3+\mathrm{five} \\ =9+6+5 \\ =20 \\ \Rightarrow p\left(3\right)=20 \\ \therefore \mathrm{The}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{polynomial}\mathrm{is}\mathrm{twenty}\mathrm{when}x=3 \end{gathered}$$ $$\begin{gathered} \left(\mathrm{iii}\right)\mathrm{To}\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}p\left(\mathrm{10}\right)\mathrm{when}x=-1,\mathrm{put}\mathrm{10}=-1\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{polynomial}. \\ p\left(x\right)={\mathrm{10}}^2+\mathrm{two}x+5 \\ \Rightarrow p\left(-1\right)={\left(-1\right)}^2+\mathrm{ii}\times \left(-1\right)+5 \\ =1-2+5 \\ =\mathrm{four} \\ \Rightarrow p\left(-\mathrm{ane}\right)=4 \\ \therefore \mathrm{The}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{polynomial}\mathrm{is}4\mathrm{when}x=-1 \end{gathered}$$ $$\begin{gathered} \left(\mathrm{iv}\right)\mathrm{To}\mathrm{discover}\mathrm{the}\mathrm{value}\mathrm{of}p\left(x\right)\mathrm{when}x=-3,\mathrm{put}x=-3\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{polynomial}. \\ p\left(x\right)={\mathrm{ten}}^2+2\mathrm{ten}+5 \\ \Rightarrow p\left(-3\right)={\left(-\mathrm{three}\right)}^{\mathrm{ii}}+2\times \left(-3\right)+5 \\ =\mathrm{nine}-\mathrm{vi}+5 \\ =8 \\ \Rightarrow p\left(-3\right)=8 \\ \therefore \mathrm{The}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{polynomial}\mathrm{is}\mathrm{viii}\mathrm{when}x=-3 \end{gathered}$$ $$\begin{gathered} \left(\mathrm{v}\right)\mathrm{To}\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}p\left(\mathrm{ten}\right)\mathrm{when}\mathrm{ten}=a,\mathrm{put}\mathrm{ten}=a\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{polynomial}. \\ p\left(x\right)={\mathrm{ten}}^{\mathrm{ii}}+2\mathrm{10}+5 \\ \Rightarrow p\left(a\right)=a^2+2\times a+5 \\ =a^2+2a+5 \\ \Rightarrow p\left(a\right)=a^2+\mathrm{ii}a+5 \\ \therefore \mathrm{The}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{polynomial}\mathrm{is}{a}^{\mathrm{two}}+2a+5\mathrm{when}x=a \end{gathered}$$

Page No lx:

Question ii:

Find the value of the polynomial y ³

$-$

5y

$-$

twoy ⁱⁱ + 3 when.

(i) y = ane
(two) y = 2
(iii) y =

$-$

2
(iv) y = four
(v) y =

$-$

b

Answer:

Permit p(y) =

$y^3-5y-2y^2+3$ $$\begin{gathered} \left(\mathrm{i}\right)\mathrm{To}\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}p\left(y\right)\mathrm{when}y=\mathrm{one},\mathrm{put}y=1\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{polynomial}. \\ p\left(y\right)=y^{\mathrm{iii}}-5y-\mathrm{ii}{y}^2+3 \\ \Rightarrow p\left(1\right)={\mathrm{i}}^{\mathrm{three}}-\mathrm{v}\times \mathrm{i}-\mathrm{ii}\times 1^{\mathrm{two}}+3 \\ =\mathrm{i}-\mathrm{v}-2+\mathrm{three} \\ =-\mathrm{three} \\ \Rightarrow p\left(\mathrm{ane}\right)=-3 \\ \therefore \mathrm{The}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{polynomial}\mathrm{is}-\mathrm{three}\mathrm{when}y=1 \end{gathered}$$ $$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{To}\mathrm{discover}\mathrm{the}\mathrm{value}\mathrm{of}p\left(y\right)\mathrm{when}y=2,\mathrm{put}y=2\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{polynomial}. \\ p\left(y\right)=y^3-5y-2y^{\mathrm{two}}+\mathrm{three} \\ \Rightarrow p\left(2\right)=2^3-\mathrm{five}\times 2-2\times 2^2+\mathrm{iii} \\ =8-\mathrm{x}-\mathrm{eight}+\mathrm{iii} \\ =-7 \\ \Rightarrow p\left(2\right)=-7 \\ \therefore \mathrm{The}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{polynomial}\mathrm{is}-7\mathrm{when}y=\mathrm{ii} \end{gathered}$$ $$\begin{gathered} \left(\mathrm{iii}\right)\mathrm{To}\mathrm{detect}\mathrm{the}\mathrm{value}\mathrm{of}p\left(y\right)\mathrm{when}y=-2,\mathrm{put}y=-2\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{polynomial}. \\ p\left(y\right)=y^3-5y-\mathrm{ii}{y}^2+\mathrm{iii} \\ \Rightarrow p\left(-2\right)={\left(-2\right)}^{\mathrm{iii}}-5\times \left(-\mathrm{ii}\right)-\mathrm{two}\times {\left(-2\right)}^2+3 \\ =-8+10-8+\mathrm{three} \\ =-3 \\ \Rightarrow p\left(-\mathrm{two}\right)=-\mathrm{iii} \\ \therefore \mathrm{The}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{polynomial}\mathrm{is}-3\mathrm{when}y=-2 \end{gathered}$$ $$\begin{gathered} \left(\mathrm{iv}\right)\mathrm{To}\mathrm{observe}\mathrm{the}\mathrm{value}\mathrm{of}p\left(y\right)\mathrm{when}y=\mathrm{iv},\mathrm{put}y=\mathrm{iv}\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{polynomial}. \\ p\left(y\right)=y^{\mathrm{three}}-\mathrm{v}y-2y^2+3 \\ \Rightarrow p\left(\mathrm{four}\right)=4^3-5\times \mathrm{iv}-\mathrm{two}\times 4^{\mathrm{two}}+\mathrm{three} \\ =64-20-32+3 \\ =15 \\ \Rightarrow p\left(\mathrm{four}\right)=15 \\ \therefore \mathrm{The}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{polynomial}\mathrm{is}15\mathrm{when}y=4 \end{gathered}$$ $$\begin{gathered} \left(\mathrm{v}\right)\mathrm{To}\mathrm{detect}\mathrm{the}\mathrm{value}\mathrm{of}p\left(y\right)\mathrm{when}y=-b,\mathrm{put}y=-b\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{polynomial}. \\ p\left(y\right)=y^3-\mathrm{five}y-2y^{\mathrm{ii}}+3 \\ \Rightarrow p\left(-b\right)={\left(-b\right)}^{\mathrm{iii}}-\mathrm{five}\times \left(-b\right)-2\times {\left(-b\right)}^{\mathrm{ii}}+\mathrm{three} \\ =-b^{\mathrm{iii}}+\mathrm{five}b-\mathrm{ii}{b}^{\mathrm{two}}+\mathrm{iii} \\ =-b^3-\mathrm{two}{b}^{\mathrm{ii}}+5b+\mathrm{iii} \\ \Rightarrow p\left(-b\right)=-b^3-2b^{\mathrm{two}}+5b+3 \\ \therefore \mathrm{The}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{polynomial}\mathrm{is}-b^3-2b^2+5b+\mathrm{three}\mathrm{when}y=-b \end{gathered}$$

Page No 60:

Question 3:

If the value of the polynomial x ² $-$ mx + 7 is 35 when x = 2 then find grand.

Answer:

$$\begin{gathered} \mathrm{Let}p\left(x\right)=x^{\mathrm{two}}-m\mathrm{ten}+\mathrm{vii} \\ \mathrm{Then}p\left(2\right)=2^2-\mathrm{grand}\times 2+\mathrm{seven} \\ =\mathrm{four}-2\mathrm{chiliad}+7 \\ \therefore p\left(\mathrm{ii}\right)=-2m+\mathrm{eleven} \\ \mathrm{Simply}p\left(2\right)=35\left(\mathrm{Given}\right) \\ \mathrm{i}.\mathrm{e}.,-2m+11=35 \\ \Rightarrow -2g=35-\mathrm{eleven} \\ \Rightarrow -2m=24 \\ \therefore k=-12 \end{gathered}$$

Page No lx:

Question 4:

The value of the polynomial ay ⁱⁱ + 2y $-$ 6 for y = $-$ three is xv, detect a.

Answer:

$$\begin{gathered} \mathrm{Allow}p\left(y\right)=a{y}^{\mathrm{two}}+2y-6 \\ \mathrm{So}p\left(-\mathrm{three}\right)=a{\left(-\mathrm{iii}\right)}^{\mathrm{two}}+\mathrm{ii}\left(-3\right)-\mathrm{half\; dozen} \\ =9a-6-6 \\ \therefore p\left(-3\right)=9a-12 \\ \mathrm{But}p\left(-\mathrm{three}\right)=15\left(\mathrm{Given}\right) \\ \mathrm{i}.\mathrm{e}.,\mathrm{nine}a-12=\mathrm{fifteen} \\ \Rightarrow 9a=15+12 \\ \Rightarrow 9a=27 \\ \therefore a=3 \end{gathered}$$

Page No 63:

Question 1:

Find the nil of the polynomial in each of the following:

(i) p(x)  = x + ii
(ii) q(10)  =  fourten

$-$

12
(iii) r(x) = 5

$-$

half-dozenx
(4) p(y)  = y + 1
(v) p(m) = thou
(vi) q(y) = foury

Respond:

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{To}\mathrm{find}\mathrm{the}\mathrm{aught}\mathrm{of}p\left(\mathrm{ten}\right),\mathrm{we}\mathrm{volition}\mathrm{solve}\mathrm{the}\mathrm{equation}p\left(x\right)=0. \\ \mathrm{We}\mathrm{have}x+2=0 \\ \Rightarrow \mathrm{ten}=-\mathrm{ii} \\ \therefore -2\mathrm{is}\mathrm{the}\mathrm{zilch}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{polynomial}. \end{gathered}$$ $$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{To}\mathrm{discover}\mathrm{the}\mathrm{zero}\mathrm{of}q\left(x\right),\mathrm{we}\mathrm{will}\mathrm{solve}\mathrm{the}\mathrm{equation}q\left(\mathrm{10}\right)=0. \\ \mathrm{We}\mathrm{have}4x-12=0 \\ \Rightarrow \mathrm{iv}\mathrm{ten}=12 \\ \Rightarrow x=3 \\ \therefore 3\mathrm{is}\mathrm{the}\mathrm{zero}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{polynomial}. \end{gathered}$$ $$\begin{gathered} \left(\mathrm{three}\right)\mathrm{To}\mathrm{notice}\mathrm{the}\mathrm{goose\; egg}\mathrm{of}r\left(\mathrm{ten}\right),\mathrm{we}\mathrm{will}\mathrm{solve}\mathrm{the}\mathrm{equation}r\left(\mathrm{10}\right)=0. \\ \mathrm{We}\mathrm{have}5-\mathrm{half-dozen}x=0 \\ \Rightarrow -6\mathrm{ten}=-5 \\ \Rightarrow x=\frac{-5}{-6}=\frac{\mathrm{v}}{6} \\ \therefore \frac{5}{\mathrm{half\; dozen}}\mathrm{is}\mathrm{the}\mathrm{zero}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{polynomial}. \end{gathered}$$ $$\begin{gathered} \left(\mathrm{iv}\right)\mathrm{To}\mathrm{find}\mathrm{the}\mathrm{zero}\mathrm{of}p\left(y\right),\mathrm{we}\mathrm{volition}\mathrm{solve}\mathrm{the}\mathrm{equation}p\left(y\right)=0. \\ \mathrm{Nosotros}\mathrm{have}y+\mathrm{ane}=0 \\ \Rightarrow y=-1 \\ \therefore -1\mathrm{is}\mathrm{the}\mathrm{zero}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{polynomial}. \end{gathered}$$ $$\begin{gathered} \left(\mathrm{v}\right)\mathrm{To}\mathrm{detect}\mathrm{the}\mathrm{zero}\mathrm{of}p\left(m\right),\mathrm{we}\mathrm{will}\mathrm{solve}\mathrm{the}\mathrm{equation}p\left(m\right)=0. \\ \mathrm{We}\mathrm{have}\mathrm{thou}=0 \\ \therefore 0\mathrm{is}\mathrm{the}\mathrm{null}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{polynomial}. \end{gathered}$$ $$\begin{gathered} \left(\mathrm{vi}\right)\mathrm{To}\mathrm{observe}\mathrm{the}\mathrm{zero}\mathrm{of}q\left(y\right),\mathrm{we}\mathrm{will}\mathrm{solve}\mathrm{the}\mathrm{equation}q\left(y\right)=0. \\ \mathrm{Nosotros}\mathrm{have}\mathrm{four}y=0 \\ \Rightarrow y=0 \\ \therefore 0\mathrm{is}\mathrm{the}\mathrm{zero}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{polynomial}. \end{gathered}$$

Page No 63:

Question 2:

Verify that:
(i) ii is a zero of the polynomial p(ten) = (x $-$ 2).
(two) 2 and ix are zeroes of the polynomial p(x) = (x $-$ 2) (x $-$ 9).
(iii) four and $-$ 3 are zeroes of the polynomial p(10) = 10 ² $-$ ten $-$ 12.

Answer:

$$\begin{gathered} \left(\mathrm{i}\right) \\ \mathrm{We}\mathrm{have}p\left(\mathrm{10}\right)=x-2 \\ \Rightarrow p\left(2\right)=\mathrm{ii}-2 \\ \Rightarrow p\left(2\right)=0 \\ \therefore 2\mathrm{is}\mathrm{a}\mathrm{zero}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{polynomial}. \end{gathered}$$ $$\begin{gathered} \left(\mathrm{ii}\right) \\ \mathrm{Nosotros}\mathrm{have}p\left(\mathrm{ten}\right)=\left(x-2\right)\left(x-9\right) \\ \Rightarrow p\left(2\right)=\left(\mathrm{two}-2\right)\left(2-9\right) \\ \Rightarrow p\left(2\right)=0\times \left(-\mathrm{seven}\right) \\ \Rightarrow p\left(\mathrm{two}\right)=0 \\ \mathrm{and}p\left(\mathrm{nine}\right)=\left(9-2\right)\left(\mathrm{nine}-9\right) \\ \Rightarrow p\left(\mathrm{ix}\right)=\mathrm{vii}\times 0 \\ \Rightarrow p\left(9\right)=0 \\ \therefore \mathrm{ii}\mathrm{and}9\mathrm{are}\mathrm{zero}e\mathrm{south}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{polynomial}. \end{gathered}$$ $$\begin{gathered} \left(\mathrm{iii}\right) \\ \mathrm{We}\mathrm{have}p\left(x\right)={\mathrm{10}}^{\mathrm{ii}}-x-12 \\ \Rightarrow p\left(4\right)={\mathrm{iv}}^2-4-12 \\ \Rightarrow p\left(4\right)=\mathrm{sixteen}-4-12 \\ \Rightarrow p\left(\mathrm{iv}\right)=0 \\ \mathrm{and}p\left(-3\right)={\left(-3\right)}^2-\left(-3\right)-12 \\ \Rightarrow p\left(-3\right)=\mathrm{nine}+3-12 \\ \Rightarrow p\left(-3\right)=0 \\ \therefore 2\mathrm{and}-3\mathrm{are}\mathrm{zeroes}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{polynomial}. \end{gathered}$$

Page No 63:

Question 3:

Find the zeroes of the post-obit quadratic polynomials and verify the relationship between the zeroes and the coefficients:
(i) x ² + 10x + 16
(ii) ten ⁱⁱ $-$ 4x $-$ five

Answer:

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{We}\mathrm{take}{x}^{\mathrm{ii}}+10\mathrm{ten}+\mathrm{xvi}=x^2+\mathrm{viii}x+\mathrm{ii}\mathrm{ten}+16 \\ =x\left(x+8\right)+\mathrm{ii}\left(\mathrm{10}+8\right) \\ =\left(x+\mathrm{eight}\right)\left(x+\mathrm{ii}\right) \\ \mathrm{The}\mathrm{polynomial}\mathrm{has}\mathrm{zeroes}\mathrm{when}p\left(x\right)=0 \\ \mathrm{i}.\mathrm{e}.,\mathrm{ten}+\mathrm{viii}=0\mathrm{or}x+2=0 \\ \Rightarrow x=-8\mathrm{or}\mathrm{ten}=-2 \\ \therefore \mathrm{The}\mathrm{zeroes}\mathrm{of}\mathrm{the}\mathrm{polynomial}{\mathrm{ten}}^2+10\mathrm{10}+16\mathrm{are}-8\mathrm{and}-2. \\ \mathrm{Now},\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{zeroes}=\left(-8\right)+\left(-2\right)=-\mathrm{x}=\frac{-\left(10\right)}{1}=\frac{-b}{a} \\ \mathrm{and}\mathrm{production}\mathrm{of}\mathrm{the}\mathrm{zeroes}=\left(-\mathrm{eight}\right)\times \left(-2\right)=16=\frac{\mathrm{sixteen}}{\mathrm{i}}=\frac{c}{a} \\ \mathrm{Hence},\mathrm{verified}. \end{gathered}$$ $$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{We}\mathrm{have}{\mathrm{10}}^{\mathrm{ii}}-4\mathrm{10}-\mathrm{v}={\mathrm{10}}^2-5x+x-5 \\ =x\left(\mathrm{10}-5\right)+\mathrm{ane}\left(\mathrm{ten}-\mathrm{five}\right) \\ =\left(\mathrm{10}-5\right)\left(x+\mathrm{one}\right) \\ \mathrm{The}\mathrm{polynomial}\mathrm{has}\mathrm{zeroes}\mathrm{when}p\left(\mathrm{10}\right)=0 \\ \mathrm{i}.\mathrm{e}.,x-5=0\mathrm{or}x+\mathrm{ane}=0 \\ \Rightarrow x=5\mathrm{or}x=-1 \\ \therefore \mathrm{The}\mathrm{zeroes}\mathrm{of}\mathrm{the}\mathrm{polynomial}{\mathrm{ten}}^{\mathrm{ii}}+\mathrm{x}x+16\mathrm{are}\mathrm{five}\mathrm{and}-\mathrm{i}. \\ \mathrm{At\; present},\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{zeroes}=5+\left(-\mathrm{one}\right)=4=\frac{-\left(-4\right)}{1}=\frac{-b}{a} \\ \mathrm{and}\mathrm{product}\mathrm{of}\mathrm{the}\mathrm{zeroes}=5\times \left(-\mathrm{ane}\right)=-5=\frac{-5}{1}=\frac{c}{a} \\ \mathrm{Hence},\mathrm{verified}. \end{gathered}$$

Page No 63:

Question 4:

Detect a quadratic polynomial, the sum and product of whose zeroes are respectively:

(i) 5 and

$-$

l
(ii)

$-$

11 and x

Answer:

$$\begin{gathered} \left(\mathrm{i}\right) \\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{a}\mathrm{quadratic}\mathrm{polynomial}\mathrm{is}\mathrm{of}\mathrm{the}\mathrm{form}{x}^2-\left(\alpha +\beta \right)x+\alpha \beta . \\ \mathrm{Hither},\alpha +\beta \mathrm{is}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{zeroes}\mathrm{and}\alpha \beta \mathrm{is}\mathrm{the}\mathrm{product}\mathrm{of}\mathrm{zeroes}. \\ \mathrm{Given}:\alpha +\beta =5\mathrm{and}\alpha \beta =-50 \\ \therefore \mathrm{Required}\mathrm{quadratic}\mathrm{polynomial}={\mathrm{10}}^2-\left(\mathrm{five}\right)\mathrm{ten}+\left(-50\right)={\mathrm{ten}}^{\mathrm{two}}-\mathrm{v}x-\mathrm{l} \end{gathered}$$ $$\begin{gathered} \left(\mathrm{two}\right)\mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{a}\mathrm{quadratic}\mathrm{polynomial}\mathrm{is}\mathrm{of}\mathrm{the}\mathrm{form}{x}^{\mathrm{ii}}-\left(\alpha +\beta \right)x+\alpha \beta . \\ \mathrm{Here},\alpha +\beta \mathrm{is}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{zeroes}\mathrm{and}\alpha \beta \mathrm{is}\mathrm{the}\mathrm{product}\mathrm{of}\mathrm{zeroes}. \\ \mathrm{Given}:\alpha +\beta =-\mathrm{xi}\mathrm{and}\alpha \beta =10 \\ \therefore \mathrm{Required}\mathrm{quadratic}\mathrm{polynomial}=x^2-\left(-11\right)\mathrm{ten}+\mathrm{ten}={\mathrm{10}}^{\mathrm{two}}+\mathrm{xi}x+\mathrm{ten} \end{gathered}$$

Folio No 64:

Question 1:

Using remainder theorem, Find the remainder when:

(i) threex ² + ten + seven is divided by x + 2.
(ii) 4x ⁱⁱⁱ + 5x

$-$

10 is divided by ten

$-$

3.
(iii) x ³

$-$

ax ² + 2x

$-$

a is divided by 10

$-$

a

Answer:

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{Let},p\left(\mathrm{10}\right)=3x^{\mathrm{two}}+\mathrm{ten}+7 \\ \mathrm{Here},\mathrm{divisor}=x+2 \\ \mathrm{When}x+2=0 \\ \Rightarrow x=-\mathrm{two} \\ \mathrm{Past}\mathrm{remainder}\mathrm{theorem}\mathrm{putting}x=-\mathrm{ii}\mathrm{in}p\left(x\right)\mathrm{we}\mathrm{get}, \\ \mathrm{Remainder}=p\left(-2\right)=3{\left(-2\right)}^2+\left(-2\right)+\mathrm{seven} \\ =\mathrm{three}\times 4-2+\mathrm{vii} \\ =12-\mathrm{ii}+7 \\ =17 \\ \therefore \mathrm{Balance}=17 \end{gathered}$$ $$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{Let}p\left(x\right)=\mathrm{four}{\mathrm{ten}}^3+\mathrm{five}x-10 \\ \mathrm{Here},\mathrm{divisor}=x-3 \\ \mathrm{When}x-3=0,\mathrm{nosotros}\mathrm{accept}\mathrm{10}=3 \\ \mathrm{Putting}x=\mathrm{three}\mathrm{in}p\left(\mathrm{ten}\right),\mathrm{we}\mathrm{get}: \\ p\left(\mathrm{three}\right)=4{\left(3\right)}^{\mathrm{three}}+5\left(3\right)-\mathrm{x} \\ =4\times 27+15-10 \\ =108+15-\mathrm{x} \\ =113 \\ \therefore \mathrm{Remainder}=113 \end{gathered}$$ $$\begin{gathered} \left(3\right)\mathrm{Let}p\left(\mathrm{ten}\right)=x^3-a{x}^2+2x-a \\ \mathrm{Hither},\mathrm{divisor}=\mathrm{10}-a \\ \mathrm{When}\mathrm{10}-a=0,\mathrm{nosotros}\mathrm{take}\mathrm{10}=a \\ \mathrm{Putting}\mathrm{ten}=a\mathrm{in}p\left(x\right),\mathrm{we}\mathrm{get}: \\ p\left(a\right)={\left(a\right)}^{\mathrm{three}}-a{\left(a\right)}^2+2\left(a\right)-a \\ =a^3-a\times a^2+2a-a \\ =a^3-a^{\mathrm{iii}}+\mathrm{two}a-a \\ =a \\ \therefore \mathrm{Remainder}=a \end{gathered}$$

Page No 64:

Question ii:

If p(x) = 210 ⁱⁱⁱ

$-$

3x ² + fourx

$-$

v. Find the balance when p(x) is divided by.

(i) 10

$-$

two
(ii) x + iii
(iii) ten

$-$

i

Respond:

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{Nosotros}\mathrm{have}p\left(\mathrm{10}\right)=2x^3-3x^2+4\mathrm{ten}-5 \\ \mathrm{Here},\mathrm{divisor}=x-\mathrm{two} \\ \mathrm{When}x-\mathrm{two}=0,\mathrm{we}\mathrm{have}x=2 \\ \mathrm{Putting}x=\mathrm{ii}\mathrm{in}p\left(x\right),\mathrm{we}\mathrm{get}: \\ p\left(2\right)=2{\left(\mathrm{ii}\right)}^3-\mathrm{iii}{\left(2\right)}^2+\mathrm{iv}\left(2\right)-5 \\ =\mathrm{two}\times 8-3\times 4+8-5 \\ =16-12+\mathrm{eight}-5 \\ =7 \\ \therefore \mathrm{Remainder}=7 \end{gathered}$$ $$\begin{gathered} \left(2\right)\mathrm{We}\mathrm{have}p\left(x\right)=2x^3-\mathrm{three}{\mathrm{10}}^2+4\mathrm{10}-5 \\ \mathrm{Here},\mathrm{divisor}=x+3 \\ \mathrm{When}\mathrm{10}+3=0,\mathrm{we}\mathrm{take}x=-3 \\ \mathrm{Putting}\mathrm{ten}=-3\mathrm{in}p\left(x\right),\mathrm{we}\mathrm{get}: \\ p\left(-3\right)=2{\left(-\mathrm{iii}\right)}^3-3{\left(-3\right)}^2+4\left(-3\right)-5 \\ =2\times \left(-27\right)-3\times \mathrm{ix}-12-\mathrm{v} \\ =-54-27-12-5 \\ =-98 \\ \therefore \mathrm{Remainder}=-98 \end{gathered}$$ $$\begin{gathered} \left(\mathrm{iii}\right)\mathrm{We}\mathrm{have}p\left(x\right)=\mathrm{two}{x}^{\mathrm{iii}}-3{\mathrm{10}}^{\mathrm{two}}+\mathrm{iv}x-\mathrm{v} \\ \mathrm{Here},\mathrm{divisor}=x-1 \\ \mathrm{When}\mathrm{ten}-1=0,\mathrm{nosotros}\mathrm{take}x=1 \\ \mathrm{Putting}x=1\mathrm{in}p\left(x\right),\mathrm{we}\mathrm{get}: \\ p\left(\mathrm{one}\right)=2{\left(1\right)}^3-3{\left(\mathrm{one}\right)}^2+\mathrm{iv}\left(1\right)-5 \\ =\mathrm{two}\times 1-\mathrm{iii}\times \mathrm{ane}+\mathrm{iv}-5 \\ =\mathrm{two}-3+4-5 \\ =-2 \\ \therefore \mathrm{Remainder}=-2 \end{gathered}$$

Page No 64:

Question iii:

When x ³ + aten ² + 4x $-$ v is divided by x + 1, the remainder is fourteen. Observe a.

Answer:

$$\begin{gathered} \mathrm{Let}p\left(x\right)=x^3+a{x}^{\mathrm{ii}}+4\mathrm{ten}-\mathrm{five} \\ \mathrm{Here},\mathrm{divisor}=\mathrm{ten}+1 \\ \mathrm{When}\mathrm{ten}+1=0,\mathrm{we}\mathrm{have}x=-1 \\ \mathrm{Putting}x=-\mathrm{i}\mathrm{in}p\left(x\right),\mathrm{we}\mathrm{become}: \\ \mathrm{Balance}=p\left(-1\right)={\left(-1\right)}^3+a{\left(-1\right)}^2+\mathrm{iv}\left(-1\right)-\mathrm{v} \\ =-\mathrm{i}+a-\mathrm{four}-\mathrm{v} \\ =a-\mathrm{ten} \\ \mathrm{Given}:\mathrm{Remainder}=14 \\ \mathrm{i}.\mathrm{e}.,a-10=\mathrm{xiv} \\ \Rightarrow a=\mathrm{xiv}+10 \\ \therefore a=24 \end{gathered}$$

Page No 65:

Question 1:

Using factor theorem , prove that:

(i) (x + ii) is gene of 10 ²

$-$

4.
(ii) (x

$-$

3) is cistron of x ⁱⁱⁱ

$-$

27.
(3) (x

$-$

1) is factor of 2x ⁴ + 9x ^three + half-dozenx ²

$-$

xix

$-$

half dozen.
(iv) (x + four) is a gene of ten ^two + 10x + 24.

Reply:

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{Let}p\left(\mathrm{10}\right)={\mathrm{ten}}^2-4 \\ \mathrm{When}x+2=0,\mathrm{we}\mathrm{have}x=-\mathrm{two} \\ \mathrm{Putting}\mathrm{10}=-2\mathrm{in}p\left(x\right),\mathrm{nosotros}\mathrm{become}: \\ p\left(-2\right)={\left(-\mathrm{ii}\right)}^2-4 \\ =4-\mathrm{four} \\ =0 \\ \therefore \mathrm{By}\mathrm{factor}\mathrm{theorem},\left(x+2\right)\mathrm{is}\mathrm{a}\mathrm{gene}\mathrm{of}{\mathrm{ten}}^{\mathrm{ii}}-\mathrm{iv}. \end{gathered}$$ $$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{Let},p\left(x\right)=x^3-27 \\ \mathrm{When}x-\mathrm{three}=0,\mathrm{we}\mathrm{have}x=3 \\ \mathrm{Putting}\mathrm{10}=\mathrm{three}\mathrm{in}p\left(x\right),\mathrm{we}\mathrm{get}: \\ p\left(3\right)={\left(3\right)}^3-27 \\ =27-27 \\ =0 \\ \therefore \mathrm{By}\mathrm{factor}\mathrm{theorem},\left(x-\mathrm{iii}\right)\mathrm{is}\mathrm{a}\mathrm{factor}\mathrm{of}{\mathrm{10}}^{\mathrm{three}}-27. \end{gathered}$$ $$\begin{gathered} \left(\mathrm{three}\right)\mathrm{Allow}p\left(x\right)=\mathrm{ii}{\mathrm{ten}}^4+9x^3+6x^2-\mathrm{eleven}x-\mathrm{half-dozen} \\ \mathrm{When}x-\mathrm{i}=0,\mathrm{we}\mathrm{have}\mathrm{ten}=1 \\ \mathrm{Putting}x=\mathrm{iii}\mathrm{in}p\left(x\right),\mathrm{nosotros}\mathrm{get}: \\ p\left(1\right)=2{\left(1\right)}^4+9{\left(1\right)}^{\mathrm{three}}+6{\left(1\right)}^{\mathrm{two}}-\mathrm{xi}\left(1\right)-6 \\ =2+9+6-\mathrm{eleven}-6 \\ =0 \\ \therefore \mathrm{By}\mathrm{cistron}\mathrm{theorem},\left(x-\mathrm{one}\right)\mathrm{is}\mathrm{a}\mathrm{factor}\mathrm{of}2x^4+\mathrm{ix}{\mathrm{ten}}^{\mathrm{three}}+6x^{\mathrm{two}}-\mathrm{xi}x-6. \end{gathered}$$ $$\begin{gathered} \left(\mathrm{iv}\right)\mathrm{Let},p\left(x\right)=x^{\mathrm{ii}}+10x+24 \\ \mathrm{When}\mathrm{10}+4=0,\mathrm{nosotros}\mathrm{take}\mathrm{10}=-4 \\ \mathrm{Put}x=-\mathrm{iv}\mathrm{in}p\left(x\right)\mathrm{we}\mathrm{get}, \\ p\left(-4\right)={\left(-4\right)}^2+10\left(-4\right)+24 \\ =16-40+24 \\ =0 \\ \therefore \mathrm{By}\mathrm{cistron}\mathrm{theorem},\left(\mathrm{ten}+4\right)\mathrm{is}\mathrm{a}\mathrm{factor}\mathrm{of}{x}^2+\mathrm{x}\mathrm{10}+24. \end{gathered}$$

Page No 65:

Question 2:

Employ gene theorem to determine whether (x $-$ 2)  is a factor of x ⁱⁱⁱ $-$ iiix ² + 4x + iv.

Answer:

$$\begin{gathered} \mathrm{Let}p\left(x\right)=x^{\mathrm{three}}-\mathrm{three}{x}^{\mathrm{two}}+4x+4 \\ \mathrm{When}\mathrm{ten}-2=0,\mathrm{we}\mathrm{have}x=2 \\ \mathrm{Putting}\mathrm{ten}=2\mathrm{in}p\left(\mathrm{ten}\right),\mathrm{nosotros}\mathrm{get}: \\ p\left(2\right)={\left(\mathrm{two}\right)}^{\mathrm{iii}}-3{\left(2\right)}^2+4\left(\mathrm{ii}\right)+\mathrm{four} \\ =\mathrm{viii}-12+8+4 \\ =8 \\ \mathrm{Nosotros}\mathrm{take}p\left(2\right)\ne 0 \\ \mathrm{i}.\mathrm{e}.,2\mathrm{is}\mathrm{not}\mathrm{the}\mathrm{nil}\mathrm{of}p\left(\mathrm{10}\right). \\ \therefore \mathrm{By}\mathrm{gene}\mathrm{theorem},\left(x-\mathrm{ii}\right)\mathrm{is}\mathrm{not}\mathrm{a}\mathrm{factor}\mathrm{of}{x}^3-3x^2+4\mathrm{10}+4. \end{gathered}$$

Page No 65:

Question 3:

Observe the value of 'a' if (x $-$ 2) is a cistron of 210 ^three $-$ half-dozen10 ² + 5x + a.

Answer:

$$\begin{gathered} \mathrm{Let}p\left(x\right)=2{\mathrm{10}}^3-6x^{\mathrm{two}}+\mathrm{v}x+a \\ \mathrm{When}x-\mathrm{two}=0,\mathrm{we}\mathrm{take}\mathrm{10}=2 \\ \mathrm{Putting}\mathrm{10}=2\mathrm{in}p\left(x\right),\mathrm{we}\mathrm{get}: \\ p\left(\mathrm{two}\right)=2{\left(\mathrm{two}\right)}^3-\mathrm{six}{\left(2\right)}^2+5\left(2\right)+a \\ =16-24+10+a \\ =a+2 \\ \left(x-2\right)\mathrm{is}\mathrm{the}\mathrm{zero}\mathrm{of}p\left(\mathrm{ten}\right). \\ \mathrm{By}\mathrm{gene}\mathrm{theorem},\mathrm{nosotros}\mathrm{accept}: \\ p\left(2\right)=0 \\ \Rightarrow a+\mathrm{ii}=0 \\ \therefore a=-2 \end{gathered}$$

Page No 173:

Question 1:

Add the following expressions:

(i) 5m + 0.3n

$-$

1.2t ; 0.23thousand

$-$

ii.8t + 4northward

(ii)

$\frac{a}{2}-\frac{7b}{3}+\frac{3c}{\mathrm{iv}};\frac{\mathrm{v}a}{2}-\frac{2b}{\mathrm{iii}}-\frac{7c}{4}$

Answer:

$$\begin{gathered} \left(\mathrm{i}\right) \\ \mathrm{We}\mathrm{have}\left(5m+0.\mathrm{three}n-1.\mathrm{two}t\right)+\left(0.23\mathrm{thousand}-2.8t+4n\right) \\ =5m+0.\mathrm{three}n-1.2t+0.23m-\mathrm{two}.8t+4\mathrm{due\; north} \\ =5m+0.23m+0.3n+4\mathrm{due\; north}-\mathrm{one}.2t-\mathrm{two}.8t \\ =5.23m+4.3n-4t \end{gathered}$$ $$\begin{gathered} \left(2\right) \\ \mathrm{We}\mathrm{take}\left(\frac{a}{\mathrm{ii}}-\frac{7b}{\mathrm{three}}+\frac{\mathrm{iii}c}{\mathrm{iv}}\right)+\left(\frac{5a}{2}-\frac{2b}{3}-\frac{7c}{4}\right) \\ =\frac{a}{2}-\frac{7b}{3}+\frac{3c}{4}+\frac{5a}{2}-\frac{2b}{\mathrm{iii}}-\frac{\mathrm{vii}c}{\mathrm{four}} \\ =\frac{a}{2}+\frac{\mathrm{five}a}{2}-\frac{7b}{3}-\frac{\mathrm{ii}b}{3}+\frac{3c}{\mathrm{four}}-\frac{7c}{4} \\ =\frac{6a}{2}-\frac{\mathrm{ix}b}{\mathrm{iii}}-\frac{4c}{4} \\ =3a-3b-c \end{gathered}$$

Folio No 173:

Question 2:

Decrease the 2d expression from the first:

(i)

$\frac{\mathrm{grand}}{3}+\frac{n}{\mathrm{two}};\frac{m}{6}+\frac{\mathrm{due\; north}}{4}$

(two)

$0.07p^{\mathrm{ii}}q+3.6p{q}^2-p^{\mathrm{two}}{q}^{\mathrm{ii}}-0.003p{q}^2+2p^{\mathrm{ii}}q-8.\mathrm{five}{p}^{\mathrm{two}}{q}^2$

Answer:

$$\begin{gathered} \left(\mathrm{i}\right) \\ \mathrm{We}\mathrm{take}\left(\frac{m}{3}+\frac{\mathrm{north}}{2}\right)-\left(\frac{\mathrm{thousand}}{6}+\frac{\mathrm{north}}{4}\right) \\ =\frac{m}{\mathrm{iii}}+\frac{n}{\mathrm{ii}}-\frac{m}{6}-\frac{\mathrm{northward}}{4} \\ =\frac{m}{3}-\frac{\mathrm{chiliad}}{\mathrm{six}}+\frac{n}{2}-\frac{n}{\mathrm{four}} \\ =\frac{2\mathrm{yard}}{6}-\frac{m}{6}+\frac{2n}{4}-\frac{n}{4} \\ =\frac{\mathrm{one\; thousand}}{\mathrm{vi}}+\frac{\mathrm{north}}{4} \end{gathered}$$ $$\begin{gathered} \left(2\right) \\ \mathrm{We}\mathrm{have}\left(0.07p^{\mathrm{ii}}q+3.6p{q}^2-p^{\mathrm{ii}}{q}^{\mathrm{ii}}\right)-\left(0.003p{q}^2-2p^{2}q+8.5p^{\mathrm{two}}{q}^{\mathrm{ii}}\right) \\ =0.07p^{2}q+\mathrm{iii}.6p{q}^2-p^{\mathrm{ii}}{q}^2-0.003p{q}^{\mathrm{two}}+\mathrm{ii}{p}^{\mathrm{ii}}q-8.5p^2{q}^{\mathrm{two}} \\ =0.07p^{2}q+2p^{\mathrm{two}}q+\mathrm{three}.6p{q}^2-0.003p{q}^{\mathrm{two}}-p^2{q}^2-8.\mathrm{five}{p}^2{q}^{\mathrm{ii}} \\ =2.07p^{2}q+\mathrm{iii}.597p{q}^{\mathrm{two}}-9.5p^{\mathrm{two}}{q}^{\mathrm{ii}} \end{gathered}$$

Page No 173:

Question 3:

Simplify:

(i)

$3c(2b-2a)+\mathrm{ii}a(b+\mathrm{three}c)-\mathrm{two}b(3c+a)$

(ii)

$g\left(m^3-{\mathrm{chiliad}}^{\mathrm{ii}}+1\right)-\mathrm{two}\left(k^4-m^3-1\right)+{\mathrm{thousand}}^2\left({\mathrm{yard}}^2-m+\mathrm{two}\right)$

(iii)

$2\left(x^2+y-3\right)+7\left(x^{\mathrm{ii}}-y+2\right)-\mathrm{three}\left(y-3{\mathrm{10}}^2-1\right)$

(four)

$\left(x+y\right)\left(\mathrm{ten}-y\right)+\left(2\mathrm{ten}-y\right)\left(\mathrm{iii}x+y\right)$

(v)

$\left(\mathrm{ii}a+b\right)\left(c-2d\right)+\left(a-b\right)\left(2c+3d\right)+\mathrm{four}\left(ac+bd\right)$

Answer:

$$\begin{gathered} \left(\mathrm{i}\right) \\ \mathrm{Nosotros}\mathrm{have}3c\left(2b-2a\right)+2a\left(b+\mathrm{iii}c\right)-2b\left(\mathrm{three}c+a\right) \\ =6bc-\mathrm{half-dozen}ac+2ab+\mathrm{six}ac-6bc-2ab \\ =\mathrm{six}bc-6bc-6ac+\mathrm{six}ac+\mathrm{ii}ab-2ab \\ =0 \end{gathered}$$ $$\begin{gathered} \left(\mathrm{ii}\right) \\ \mathrm{Nosotros}\mathrm{have}\mathrm{thousand}\left(m^3-m^2+1\right)-2\left(m^4-{\mathrm{thou}}^3-\mathrm{one}\right)+m^{\mathrm{ii}}\left(k^2-m+\mathrm{ii}\right) \\ =m^{\mathrm{iv}}-m^3+g-2{\mathrm{yard}}^4+2m^3+\mathrm{ii}+g^{\mathrm{iv}}-{\mathrm{yard}}^3+2{\mathrm{one\; thousand}}^2 \\ ={\mathrm{thou}}^4-2{\mathrm{one\; thousand}}^4+m^{\mathrm{four}}-{\mathrm{chiliad}}^3+2{\mathrm{thousand}}^3-m^{\mathrm{three}}+2{\mathrm{yard}}^2+\mathrm{1000}+2 \\ =2{\mathrm{thousand}}^2+m+\mathrm{two} \end{gathered}$$ $$\begin{gathered} \left(\mathrm{iii}\right) \\ \mathrm{We}\mathrm{have}2\left(x^{\mathrm{two}}+y-3\right)+\mathrm{vii}\left(x^2-y+\mathrm{ii}\right)-3\left(y-\mathrm{three}{\mathrm{10}}^{\mathrm{ii}}-1\right) \\ =2x^2+2y-6+7x^2-7y+\mathrm{xiv}-3y+9{\mathrm{ten}}^2+3 \\ =\mathrm{ii}{\mathrm{ten}}^2+7x^2+\mathrm{nine}{x}^2+2y-7y-3y-\mathrm{vi}+14+3 \\ =18x^{\mathrm{ii}}-\mathrm{viii}y+\mathrm{xi} \end{gathered}$$ $$\begin{gathered} \left(4\right) \\ \mathrm{We}\mathrm{have}\left(x+y\right)\left(x-y\right)+\left(2x-y\right)\left(3x+y\right) \\ =x\left(x-y\right)+y\left(\mathrm{10}-y\right)+2x\left(3x+y\right)-y\left(\mathrm{three}x+y\right) \\ ={\mathrm{10}}^2-xy+xy-y^{\mathrm{two}}+6x^2+\mathrm{ii}\mathrm{ten}y-3xy-y^{\mathrm{ii}} \\ ={\mathrm{ten}}^2+6x^2-\mathrm{ten}y+\mathrm{ten}y+2xy-3\mathrm{ten}y-y^2-y^2 \\ =7x^2-xy-\mathrm{two}{y}^{\mathrm{ii}} \end{gathered}$$ $$\begin{gathered} \left(\mathrm{five}\right) \\ \mathrm{We}\mathrm{have}\left(\mathrm{two}a+b\right)\left(c-2d\right)+\left(a-b\right)\left(2c+3d\right)+4\left(ac+bd\right) \\ =\mathrm{two}a\left(c-2d\right)+b\left(c-2d\right)+a\left(\mathrm{ii}c+3d\right)-b\left(2c+3d\right)+4\left(ac+bd\right) \\ =2ac-4ad+bc-2bd+\mathrm{two}ac+\mathrm{iii}ad-2bc-3bd+\mathrm{four}ac+4bd \\ =2ac+2ac+4ac-4ad+\mathrm{three}ad+bc-\mathrm{ii}bc-\mathrm{ii}bd-3bd+4bd \\ =8ac-ad-bc-bd \end{gathered}$$

Page No 173:

Question iv.ane:

Factorise:
$\frac{x^2}{16}-\frac{y^{\mathrm{two}}}{25}$

Answer:

$$\begin{gathered} \mathrm{Nosotros}\mathrm{take}\frac{x^{\mathrm{two}}}{\mathrm{xvi}}-\frac{y^{\mathrm{two}}}{25} \\ ={\left(\frac{x}{4}\right)}^2-{\left(\frac{y}{\mathrm{five}}\right)}^{\mathrm{two}} \\ =\left(\frac{x}{4}+\frac{y}{\mathrm{five}}\right)\left(\frac{\mathrm{ten}}{\mathrm{four}}-\frac{y}{\mathrm{v}}\right) \end{gathered}$$

Page No 173:

Question 4.2:

ten ⁱⁱ $-$ (2y + 3z)ⁱⁱ

Answer:

$$\begin{gathered} \mathrm{We}\mathrm{have}{x}^{\mathrm{ii}}-{\left(2y+3z\right)}^2 \\ =\left[x+\left(2y+3z\right)\right]\left[x-\left(2y+\mathrm{three}z\right)\right] \\ =\left(\mathrm{10}+2y+\mathrm{three}z\right)\left(x-2y-3z\right) \end{gathered}$$

Page No 173:

Question 4.three:

$16{\left(a-b\right)}^2-\mathrm{iv}{\left(c-d\right)}^{\mathrm{ii}}$

Answer:

$$\begin{gathered} \mathrm{We}\mathrm{have}16{\left(a-b\right)}^2-\mathrm{four}{\left(c-d\right)}^2 \\ =\mathrm{four}\left[\mathrm{iv}{\left(a-b\right)}^{\mathrm{ii}}-{\left(c-d\right)}^2\right] \\ =\mathrm{four}\left[{\left\{2\left(a-b\right)\right\}}^2-{\left(c-d\right)}^{\mathrm{two}}\right] \\ =\mathrm{four}\left[2\left(a-b\right)+\left(c-d\right)\right]\left[2\left(a-b\right)-\left(c-d\right)\right] \\ =\mathrm{four}\left(2a-2b+c-d\right)\left(2a-\mathrm{ii}b-c+d\right) \end{gathered}$$

Folio No 173:

Question 4.4:

(a $-$ b)²  + eight(a $-$ b) + fifteen

Respond:

$$\begin{gathered} \mathrm{We}\mathrm{take}{\left(a-b\right)}^{\mathrm{two}}+8\left(a-b\right)+\mathrm{xv} \\ ={\left(a-b\right)}^2+5\left(a-b\right)+3\left(a-b\right)+\mathrm{xv} \\ =\left(a-b\right)\left[\left(a-b\right)+5\right]+3\left[\left(a-b\right)+5\right] \\ =\left[\left(a-b\right)+5\right]\left[\left(a-b\right)+3\right] \\ =\left(a-b+\mathrm{v}\right)\left(a-b+3\right) \end{gathered}$$

Folio No 173:

Question iv.five:

9a ² $-$ 18ab $-$ 4ab ² + eightb ³

Answer:

$$\begin{gathered} \mathrm{Nosotros}\mathrm{have}\mathrm{ix}{a}^2-18ab-4a{b}^2+8b^3 \\ =9a\left(a-2b\right)-4b^{\mathrm{ii}}\left(a-2b\right) \\ =\left(a-2b\right)\left(\mathrm{ix}a-4b^2\right) \end{gathered}$$

Page No 173:

Question 4.6:

6x ^two + fourxy $-$ 9xy ² $-$ half-dozeny ³

Respond:

$$\begin{gathered} \mathrm{We}\mathrm{accept}\mathrm{six}{x}^2+4xy-9x{y}^2-6y^{\mathrm{iii}} \\ =\mathrm{two}x\left(3x+2y\right)-\mathrm{three}{y}^{\mathrm{two}}\left(\mathrm{iii}x+\mathrm{two}y\right) \\ =\left(3x+2y\right)\left(\mathrm{two}x-\mathrm{iii}{y}^2\right) \end{gathered}$$

Folio No 173:

Question 4.7:

$a^2+4ab+4b^2-9m^2+\mathrm{vi}\mathrm{thousand}n-{\mathrm{north}}^2$

Answer:

$$\begin{gathered} \mathrm{We}\mathrm{have}{a}^2+\mathrm{iv}ab+4b^{\mathrm{ii}}-\mathrm{ix}{m}^{\mathrm{two}}+6m\mathrm{due\; north}-n^2 \\ =\left(a^2+4ab+\mathrm{four}{b}^2\right)-\left(9m^2-\mathrm{half\; dozen}m\mathrm{north}+n^{\mathrm{two}}\right) \\ =\left[{\left(a\right)}^2+\mathrm{ii}\times a\times 2b+{\left(\mathrm{ii}b\right)}^2\right]-\left[{\left(3m\right)}^2-2\times 3g\times \mathrm{northward}+{\left(n\right)}^2\right] \\ ={\left(a+2b\right)}^2-{\left(\mathrm{iii}m-n\right)}^2 \\ =\left[\left(a+2b\right)+\left(\mathrm{three}\mathrm{yard}-n\right)\right]\left[\left(a+2b\right)-\left(3m-n\right)\right] \\ =\left(a+\mathrm{two}b+3m-\mathrm{north}\right)\left(a+2b-3\mathrm{1000}+\mathrm{due\; north}\right) \end{gathered}$$

Page No 173:

Question 4.8:

(2a + b)³ $-$ (a + 3b)³

Answer:

$$\begin{gathered} \mathrm{We}\mathrm{have}{\left(2a+b\right)}^3-{\left(a+3b\right)}^3 \\ =\left[\left(\mathrm{two}a+b\right)-\left(a+\mathrm{iii}b\right)\right]\left[{\left(\mathrm{ii}a+b\right)}^{\mathrm{two}}+\left(2a+b\right)\left(a+3b\right)+{\left(a+3b\right)}^{\mathrm{two}}\right] \\ =\left(a-2b\right)\left(4a^2+b^2+\mathrm{iv}ab+\mathrm{ii}{a}^{\mathrm{two}}+7ab+3b^{\mathrm{two}}+a^2+\mathrm{ix}{b}^2+\mathrm{half-dozen}ab\right) \\ =\left(a-2b\right)\left(4a^{\mathrm{two}}+2a^2+a^2+b^2+\mathrm{iii}{b}^2+\mathrm{nine}{b}^2+4ab+7ab+6ab\right) \\ =\left(a-2b\right)\left(7a^2+13b^2+17ab\right) \end{gathered}$$

Page No 173:

Question 4.9:

$\mathrm{viii}{a}^{\mathrm{three}}+b^{\mathrm{iii}}+12a^{2}b+\mathrm{vi}a{b}^2$

Answer:

$$\begin{gathered} \mathrm{We}\mathrm{have}8a^{\mathrm{three}}+b^3+12a^{\mathrm{two}}b+6a{b}^2 \\ ={\left(\mathrm{ii}a\right)}^{\mathrm{iii}}+{\left(b\right)}^{\mathrm{three}}+6ab\left(2a+b\right) \\ ={\left(2a\right)}^3+{\left(b\right)}^{\mathrm{iii}}+\mathrm{iii}\times \mathrm{ii}a\times b\left(2a+b\right) \\ ={\left(\mathrm{ii}a+b\right)}^3 \\ =\left(2a+b\right)\left(2a+b\right)\left(2a+b\right) \end{gathered}$$

Page No 173:

Question 4.10:

$\mathrm{two}{\left({\mathrm{ten}}^2-\mathrm{half-dozen}\mathrm{10}\right)}^2-8\left(x^{\mathrm{two}}-\mathrm{six}\mathrm{10}+3\right)-\mathrm{twoscore}$

Answer:

$$\begin{gathered} \mathrm{We}\mathrm{accept}2{\left({\mathrm{ten}}^2-6x\right)}^{\mathrm{two}}-8\left(x^2-6\mathrm{10}+3\right)-40 \\ \mathrm{Substituting}\left(x^2-\mathrm{vi}x\right)=a,\mathrm{we}\mathrm{get}: \\ 2{\left(x^2-6x\right)}^2-8\left({\mathrm{ten}}^2-6x+3\right)-\mathrm{forty}=2a^2-8\left(a+3\right)-\mathrm{twoscore} \\ =2a^{\mathrm{ii}}-\mathrm{viii}a-24-40 \\ =2a^2-8a-64 \\ =2\left(a^2-4a-32\right) \\ =\mathrm{two}\left(a^2-8a+4a-32\right) \\ =\mathrm{two}\left[a\left(a-8\right)+\mathrm{iv}\left(a-8\right)\right] \\ =\mathrm{ii}\left(a-8\right)\left(a+4\right) \\ \mathrm{Resubstituting}\mathrm{the}\mathrm{value}\mathrm{of}a\mathit{,}\mathrm{nosotros}\mathrm{go}: \\ \mathrm{ii}{\left(x^2-6x\right)}^{\mathrm{two}}-8\left(x^2-\mathrm{six}\mathrm{10}+\mathrm{iii}\right)-40=\mathrm{ii}\left(x^2-\mathrm{six}\mathrm{10}-8\right)\left(x^2-\mathrm{six}x+4\right) \end{gathered}$$

Page No 173:

Question 4.11:

$\left(a^2-\mathrm{two}a+3\right)\left(a^2-\mathrm{two}a+5\right)-35$

Respond:

$$\begin{gathered} \mathrm{We}\mathrm{have}\left(a^2-\mathrm{ii}a+3\right)\left(a^2-2a+\mathrm{v}\right)-35 \\ \mathrm{Substituting}\left(a^2-2a\right)=x,\mathrm{we}\mathrm{get}: \\ \left(a^{\mathrm{two}}-\mathrm{ii}a+3\right)\left(a^{\mathrm{ii}}-\mathrm{two}a+\mathrm{five}\right)-35=\left(\mathrm{10}+3\right)\left(x+\mathrm{five}\right)-35 \\ ={\mathrm{10}}^2+5x+\mathrm{iii}x+15-35 \\ =x^2+8x-20 \\ =x^2+10\mathrm{ten}-\mathrm{ii}x-20 \\ =x\left(x+10\right)-2\left(\mathrm{ten}+10\right) \\ =\left(x+10\right)\left(\mathrm{10}-\mathrm{ii}\right) \\ \mathrm{Resubstituting}\mathrm{the}\mathrm{value}\mathrm{of}a\mathit{,}\mathrm{we}\mathrm{get}: \\ \left(a^2-2a+3\right)\left(a^{\mathrm{ii}}-2a+\mathrm{five}\right)-35=\left(a^2-2a+10\right)\left(a^{\mathrm{ii}}-\mathrm{two}a-2\right) \end{gathered}$$

Page No 173:

Question 4.12:

$x^3-\mathrm{viii}{y}^3-64z^3-24\mathrm{10}yz$

Reply:

$$\begin{gathered} \mathrm{We}\mathrm{have}{x}^3-\mathrm{eight}{y}^{\mathrm{iii}}-64z^{\mathrm{iii}}-24xyz \\ ={\left(\mathrm{10}\right)}^{\mathrm{three}}+{\left(-\mathrm{ii}y\right)}^{\mathrm{iii}}+{\left(-4z\right)}^3-\mathrm{three}\times \left(\mathrm{10}\right)\times \left(-\mathrm{two}y\right)\times \left(-4z\right) \\ =\left[\left(x\right)+\left(-2y\right)+\left(-4z\right)\right]\left[{\left(\mathrm{10}\right)}^{\mathrm{two}}+{\left(-2y\right)}^2+{\left(-4z\right)}^2-\left(x\right)\left(-2y\right)-\left(-2y\right)\left(-4z\right)-\left(-4z\right)\left(x\right)\right] \\ =\left(x-\mathrm{two}y-\mathrm{iv}z\right)\left({\mathrm{ten}}^2+4y^{\mathrm{ii}}+\mathrm{xvi}{z}^{\mathrm{ii}}+2xy-8yz+4xz\right) \end{gathered}$$

Page No 173:

Question 4.13:

$p^{\mathrm{iii}}-10+\frac{27}{p^{\mathrm{three}}}$

Answer:

$$\begin{gathered} \mathrm{We}\mathrm{have}{p}^3-10+\frac{27}{p^3} \\ =p^3+\mathrm{eight}-18+\frac{27}{p^{\mathrm{three}}} \\ =p^3+8+\frac{27}{p^{\mathrm{three}}}-18 \\ ={\left(p\right)}^{\mathrm{three}}+{\left(\mathrm{ii}\right)}^{\mathrm{three}}+{\left(\frac{3}{p}\right)}^3-3\times \left(p\right)\times \left(2\right)\times \left(\frac{\mathrm{iii}}{p}\right) \\ =\left(p+2+\frac{\mathrm{iii}}{p}\right)\left[p^2+2^{\mathrm{ii}}+{\left(\frac{3}{p}\right)}^2-p\times 2-2\times \frac{3}{p}-\frac{3}{p}\times p\right] \\ =\left(p+2+\frac{\mathrm{iii}}{p}\right)\left[p^{\mathrm{ii}}+4+\frac{9}{p^2}-\mathrm{ii}p-\frac{6}{p}-\mathrm{three}\right] \end{gathered}$$

Folio No 173:

Question four.14:

$8x^6+95x^3+1$

Answer:

$$\begin{gathered} \mathrm{We}\mathrm{have}8{\mathrm{10}}^6+95x^{\mathrm{three}}+\mathrm{one} \\ =\mathrm{eight}{\mathrm{ten}}^{\mathrm{half\; dozen}}+125x^3-30x^3+\mathrm{one} \\ =8{\mathrm{10}}^{\mathrm{six}}+125x^3+1-\mathrm{xxx}{\mathrm{10}}^3 \\ ={\left(\mathrm{ii}{\mathrm{ten}}^2\right)}^3+{\left(5x\right)}^3+{\left(1\right)}^{\mathrm{iii}}-3\times \left(2{\mathrm{10}}^2\right)\times \left(\mathrm{v}\mathrm{ten}\right)\times \left(\mathrm{i}\right) \\ =\left(2x^{\mathrm{ii}}+5x+1\right)\left[{\left(\mathrm{two}{x}^{\mathrm{ii}}\right)}^{\mathrm{ii}}+{\left(5\mathrm{10}\right)}^2+{\left(\mathrm{ane}\right)}^{\mathrm{two}}-\left(2{\mathrm{ten}}^2\right)\times \left(5x\right)-\left(5x\right)\times \left(\mathrm{one}\right)-\left(1\right)\times \left(\mathrm{ii}{x}^2\right)\right] \\ =\left(2x^2+\mathrm{v}x+1\right)\left(4x^4+25x^2+1-\mathrm{x}{x}^{\mathrm{iii}}-5x-2x^{\mathrm{ii}}\right) \\ =\left(\mathrm{ii}{\mathrm{10}}^{\mathrm{ii}}+\mathrm{v}\mathrm{ten}+\mathrm{one}\right)\left(4x^4-10{\mathrm{10}}^{\mathrm{iii}}+25x^2-2{\mathrm{ten}}^2-5x+1\right) \\ =\left(2{\mathrm{ten}}^2+5x+1\right)\left(4x^4-10{\mathrm{ten}}^3+23x^2-\mathrm{five}\mathrm{10}+\mathrm{i}\right) \end{gathered}$$

Page No 173:

Question 4.15:

$\mathrm{eight}{x}^3+27y^{\mathrm{three}}+125z^3-\mathrm{ninety}xyz$

Reply:

$$\begin{gathered} \mathrm{We}\mathrm{have}\mathrm{eight}{\mathrm{10}}^{\mathrm{iii}}+27y^3+125z^{\mathrm{iii}}-\mathrm{xc}\mathrm{ten}yz \\ ={\left(2x\right)}^3+{\left(\mathrm{three}y\right)}^3+{\left(\mathrm{five}z\right)}^3-\mathrm{three}\times \left(2x\right)\times \left(\mathrm{three}y\right)\times \left(5z\right) \\ =\left(\mathrm{ii}x+3y+5z\right)\left[{\left(\mathrm{ii}x\right)}^{\mathrm{two}}+{\left(3y\right)}^{\mathrm{two}}+{\left(5z\right)}^2-\left(\mathrm{two}\mathrm{ten}\right)\times \left(3y\right)-\left(3y\right)\times \left(5z\right)-\left(5z\right)\times \left(2x\right)\right] \\ =\left(2x+3y+\mathrm{five}z\right)\left(4x^2+\mathrm{ix}{y}^2+25z^2-\mathrm{half\; dozen}xy-15yz-\mathrm{ten}xz\right) \end{gathered}$$

Page No 173:

Question 5:

What should exist added to $6x^2-3xy+4y^2$ to go $2y^2+xy-\mathrm{iv}{x}^{\mathrm{ii}}$?

Respond:

$$\begin{gathered} \mathrm{Let}p\left(x\right)\mathrm{exist}\mathrm{added}\mathrm{to}\mathrm{half\; dozen}{\mathrm{ten}}^2-\mathrm{iii}xy+\mathrm{four}{y}^2\mathrm{to}\mathrm{get}2y^{\mathrm{ii}}+xy-\mathrm{iv}{x}^2. \\ \mathrm{i}.\mathrm{e}.,\mathrm{half-dozen}{\mathrm{ten}}^{\mathrm{two}}-\mathrm{three}xy+4y^2+p\left(x\right)=2y^{\mathrm{ii}}+xy-\mathrm{iv}{x}^2 \\ \Rightarrow p\left(x\right)=\left(2y^2+\mathrm{10}y-4{\mathrm{10}}^2\right)-\left(6x^{\mathrm{two}}-3xy+\mathrm{iv}{y}^2\right) \\ \Rightarrow p\left(\mathrm{ten}\right)=\mathrm{ii}{y}^2+xy-\mathrm{iv}{\mathrm{ten}}^{\mathrm{two}}-6x^2+3\mathrm{ten}y-4y^2 \\ \Rightarrow p\left(\mathrm{10}\right)=2y^2-\mathrm{iv}{y}^2+xy+\mathrm{iii}\mathrm{ten}y-4{\mathrm{10}}^{\mathrm{two}}-6x^2 \\ \Rightarrow p\left(x\right)=-2y^2+\mathrm{iv}\mathrm{10}y-10x^2 \\ \Rightarrow p\left(x\right)=-\mathrm{x}{\mathrm{ten}}^2-2y^2+4xy \\ \therefore -10x^2-2y^2+4xy\mathrm{tin\; can}\mathrm{be}\mathrm{added}\mathrm{to}\mathrm{half\; dozen}{x}^2-3\mathrm{10}y+4y^2\mathrm{to}\mathrm{go}2y^{\mathrm{two}}+xy-4x^2. \end{gathered}$$

Page No 173:

Question half-dozen:

What should be subtracted from 31000 ² due north + 5mn ² $-$ iiyard ² north ^two to go $-$3m ^two n + 7mn ² + 4m ⁱⁱ n ² ?

Respond:

$$\begin{gathered} \mathrm{Let}p\left(\mathrm{10}\right)\mathrm{exist}\mathrm{subtracted}\mathrm{from}\mathrm{three}{m}^{2}n+5m{n}^{\mathrm{ii}}-2m^{\mathrm{two}}{\mathrm{north}}^{2}to\mathrm{1000}et-3{\mathrm{chiliad}}^2\mathrm{due\; north}+7\mathrm{thousand}{\mathrm{northward}}^2+4m^2{\mathrm{due\; north}}^2. \\ \therefore \left(\mathrm{three}{\mathrm{thousand}}^2\mathrm{due\; north}+\mathrm{v}m{n}^2-2{\mathrm{thou}}^2{n}^{\mathrm{ii}}\right)-p\left(x\right)=\left(-3m^{2}n+7m{\mathrm{north}}^2+\mathrm{iv}{m}^2{\mathrm{due\; north}}^2\right) \\ \Rightarrow p\left(x\right)=\left(\mathrm{three}{m}^{2}n+5\mathrm{one\; thousand}{\mathrm{northward}}^2-2{\mathrm{yard}}^{\mathrm{ii}}{n}^2\right)-\left(-3m^{2}n+7g{n}^{\mathrm{two}}+4{\mathrm{yard}}^2{n}^{\mathrm{two}}\right) \\ \Rightarrow p\left(\mathrm{ten}\right)=\mathrm{iii}{m}^2\mathrm{northward}+5m{n}^2-2{\mathrm{one\; thousand}}^{\mathrm{ii}}{n}^2+\mathrm{iii}{\mathrm{1000}}^{\mathrm{ii}}n-7\mathrm{yard}{\mathrm{north}}^2-\mathrm{four}{\mathrm{yard}}^2{\mathrm{north}}^2 \\ \Rightarrow p\left(x\right)=\mathrm{three}{\mathrm{thousand}}^{2}n+\mathrm{iii}{m}^2\mathrm{north}+\mathrm{five}m{\mathrm{northward}}^2-7\mathrm{1000}{\mathrm{north}}^2-2m^{\mathrm{ii}}{\mathrm{north}}^{\mathrm{ii}}-\mathrm{four}{k}^2{\mathrm{due\; north}}^2 \\ \Rightarrow p\left(\mathrm{ten}\right)=6m^2\mathrm{northward}-2m{n}^2-6m^2{\mathrm{due\; north}}^2 \\ \therefore \left(\mathrm{half\; dozen}{k}^{\mathrm{ii}}\mathrm{due\; north}-2\mathrm{chiliad}{\mathrm{northward}}^2-6{\mathrm{one\; thousand}}^2{n}^2\right)\mathrm{can}\mathrm{be}\mathrm{subtracted}\mathrm{from}\left(\mathrm{iii}{\mathrm{yard}}^2\mathrm{due\; north}+\mathrm{v}\mathrm{1000}{n}^2-\mathrm{ii}{m}^{\mathrm{two}}{n}^{\mathrm{ii}}\right)\mathrm{to}\mathrm{get}\left(-3g^{2}n+7g{\mathrm{north}}^2+\mathrm{iv}{m}^2{n}^2\right). \end{gathered}$$

Page No 173:

Question 7:

Subtract vix + 8y + 4 from the lord's day of ivx $-$ twoy + 3 and 2y $-$ ten + 3?

Respond:

$$\begin{gathered} \mathrm{Required}\mathrm{polynomial}=\left(4x-\mathrm{ii}y+\mathrm{three}\right)+\left(2y-x+3\right)-\left(6x+\mathrm{eight}y+\mathrm{iv}\right) \\ =\mathrm{four}x-\mathrm{ii}y+3+2y-x+\mathrm{iii}-6x-\mathrm{eight}y-4 \\ =\mathrm{four}x-x-\mathrm{half\; dozen}\mathrm{ten}-2y+\mathrm{two}y-8y+3+3-4 \\ =-3\mathrm{ten}-\mathrm{eight}y+\mathrm{two} \end{gathered}$$

Page No 173:

Question 8:

Find the perimeter of a rectangle whose two next sides are
vten ² + 2xy $-$ xiii ; iix ⁱⁱ $-$ vixy + 11.

Answer:

$$\begin{gathered} \mathrm{Two}\mathrm{next}\mathrm{sides}\mathrm{of}\mathrm{a}\mathrm{rectangle}\mathrm{are}\mathrm{its}\mathrm{length}\mathrm{and}\mathrm{latitude}. \\ \therefore \mathrm{Perimeter}=2\left(\mathrm{length}+\mathrm{breadth}\right) \\ =\mathrm{ii}\left(\mathrm{sum}\mathrm{of}\mathrm{two}\mathrm{side\; by\; side}\mathrm{sides}\right) \\ =2\left[\left(5{\mathrm{ten}}^2+2\mathrm{10}y-13\right)+\left(\mathrm{two}{x}^2-6xy+11\right)\right] \\ =2\left(\mathrm{v}{\mathrm{ten}}^{\mathrm{ii}}+\mathrm{two}xy-13+\mathrm{two}{x}^{\mathrm{ii}}-6\mathrm{10}y+11\right) \\ =\mathrm{ii}\left(5x^2+2{\mathrm{10}}^2+2xy-6xy-13+11\right) \\ =\mathrm{two}\left(7{\mathrm{ten}}^2-\mathrm{iv}xy-2\right) \\ =\mathrm{xiv}{\mathrm{ten}}^2-8xy-4 \end{gathered}$$

Folio No 173:

Question 9:

The perimeter of a triangle is 9m ² $-$ iin + 8 and its two sides are fourm ² + 3n and 7grand ² + vnorth $-$ 12. Find the 3rd side of the triangle.

Answer:

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{showtime}\mathrm{side}\mathrm{of}\mathrm{the}\mathrm{triangle}\mathrm{exist}4{\mathrm{yard}}^2+3n. \\ \mathrm{2nd}\mathrm{side}\mathrm{of}\mathrm{triangle}=7m^2+\mathrm{v}n-12 \\ \mathrm{Third}\mathrm{side}=p\left(x\right) \\ \mathrm{Perimeter}=9m^{\mathrm{two}}-\mathrm{two}\mathrm{northward}+8 \\ \mathrm{At\; present},4{\mathrm{1000}}^2+3\mathrm{northward}+\mathrm{seven}{m}^2+5n-12+p\left(x\right)=9m^2-\mathrm{two}n+\mathrm{viii} \\ \Rightarrow p\left(x\right)=9{\mathrm{one\; thousand}}^2-\mathrm{two}\mathrm{due\; north}+\mathrm{viii}-\left(4m^2+\mathrm{three}\mathrm{due\; north}+7m^{\mathrm{ii}}+5n-12\right) \\ \Rightarrow p\left(\mathrm{10}\right)=9m^2-2n+8-\mathrm{four}{m}^2-3\mathrm{due\; north}-\mathrm{vii}{g}^2-5n+12 \\ \Rightarrow p\left(\mathrm{10}\right)=9m^2-\mathrm{iv}{m}^2-\mathrm{seven}{\mathrm{one\; thousand}}^2-2n-3n-5n+\mathrm{viii}+12 \\ \Rightarrow p\left(\mathrm{ten}\right)=-2m^2-10n+20 \\ \therefore \mathrm{3rd}\mathrm{side}\mathrm{of}\mathrm{the}\mathrm{triangle}=-2m^2-10n+20 \end{gathered}$$

Page No 173:

Question x:

Rahul'south monthly salary is Rs. twop ³ + p $-$ 3. His annual expenditure is Rs 14p ^two + 6p $-$ 10. Detect his annual savings.

Answer:

$$\begin{gathered} \mathrm{Given}:\mathrm{Rahul}\text{'}\mathrm{s}\mathrm{monthly}\mathrm{salary}=\mathrm{Rs}\left(2p^{\mathrm{two}}+p-\mathrm{three}\right) \\ A\mathrm{nnual}\mathrm{salary}=12\left(\mathrm{monthly}\mathrm{salary}\right)=\mathrm{Rs}12\left(2p^2+p-3\right) \\ =\mathrm{Rs}\left(24p^{\mathrm{ii}}+12p-36\right) \\ \mathrm{And}\mathrm{his}\mathrm{almanac}\mathrm{expenditure}=\mathrm{Rs}\left(14p^{\mathrm{ii}}+\mathrm{half-dozen}p-10\right) \\ \mathrm{Annual}\mathrm{saving}=\mathrm{Annual}\mathrm{salary}-\mathrm{almanac}\mathrm{expenditure} \\ =\mathrm{Rs}\left(24p^{\mathrm{ii}}+12p-36\right)-\mathrm{Rs}\left(\mathrm{xiv}{p}^2+6p-10\right) \\ =\mathrm{Rs}\left[\left(24p^2+12p-36\right)-\left(14p^2+6p-10\right)\right] \\ =\mathrm{Rs}\left(24p^2+12p-36-\mathrm{xiv}{p}^{\mathrm{ii}}-\mathrm{half\; dozen}p+10\right) \\ =\mathrm{Rs}\left(24p^2-14p^2+12p-\mathrm{half-dozen}p-36+10\right) \\ =\mathrm{Rs}\left(10p^2+\mathrm{half\; dozen}p-26\right) \end{gathered}$$

Page No 173:

Question 11.1:

Apply constructed partition method for performing following divisions. Write whether the divisor is a gene of dividend. Justify.

(i) (3p ⁴

$-$

ivp ³

$-$

3p

$-$

ane)

$÷$

(p

$-$

1)

Answer:

$$\begin{gathered} \mathrm{Dividend}=3p^4-\mathrm{four}{p}^3-3p-1 \\ \mathrm{Index}\mathrm{course}=3p^{\mathrm{iv}}-4p^3+0p^{\mathrm{ii}}-3p-\mathrm{one} \\ \mathrm{Coefficient}\mathrm{grade}=\left(\mathrm{iii},-4,0,-3,-1\right) \\ \mathrm{Comparison}\mathrm{the}\mathrm{divisor}p-\mathrm{one}\mathrm{with}p-a,\mathrm{we}\mathrm{get}a=1 \end{gathered}$$

$$\begin{gathered} \therefore \mathrm{Quotient}\mathrm{in}\mathrm{coefficient}\mathrm{form}=\left(\mathrm{three},-\mathrm{one},-1,-\mathrm{iv}\right) \\ \mathrm{i}.\mathrm{e}.,\mathrm{quotient}=3p^3-p^{\mathrm{two}}-p-4\mathrm{in}\mathrm{the}\mathrm{variable}p \\ \mathrm{Here},\mathrm{the}\mathrm{rest}\mathrm{is}-5\left(\mathrm{nonzero}\right). \\ \therefore \left(p-\mathrm{i}\right)\mathrm{is}\mathrm{not}\mathrm{a}\mathrm{factor}\mathrm{of}\left(3p^4-4p^3-\mathrm{iii}p-1\right). \end{gathered}$$

Folio No 173:

Question xi.2:

(ii) (x ³ $-$ 8) $÷$ (x $-$ 2)

Answer:

$$\begin{gathered} \mathrm{Dividend}=x^3-8 \\ \mathrm{Index}\mathrm{form}=x^{\mathrm{iii}}+0{\mathrm{ten}}^{\mathrm{two}}+0\mathrm{ten}-8 \\ \mathrm{Coefficient}\mathrm{form}=\left(\mathrm{ane},0,0,-8\right) \\ \mathrm{Comparing}\mathrm{the}\mathrm{divisor}x-\mathrm{two}\mathrm{with}\mathrm{10}-a,\mathrm{we}\mathrm{get}a=\mathrm{ii} \end{gathered}$$

$$\begin{gathered} \therefore \mathrm{Caliber}\mathrm{in}\mathrm{coefficient}\mathrm{form}=\left(\mathrm{one},\mathrm{two},4\right) \\ \mathrm{i}.\mathrm{e}.,\mathrm{quotient}=x^{\mathrm{two}}+2x+4\mathrm{in}\mathrm{the}\mathrm{variable}x \\ \mathrm{Here},\mathrm{the}\mathrm{remainder}\mathrm{is}0\left(\mathrm{null}\right). \\ \therefore \left(x-\mathrm{ii}\right)\mathrm{is}\mathrm{a}\mathrm{factor}\mathrm{of}\left({\mathrm{ten}}^{\mathrm{three}}-8\right). \end{gathered}$$

Page No 173:

Question 11.3:

(4x ^four + tenten ⁱⁱⁱ $-$ threeten ^two + 2x $-$ 21) $÷$ (x + iii)

Respond:

$$\begin{gathered} \mathrm{Dividend}=4x^{\mathrm{four}}+10x^3-3x^2+2\mathrm{10}-21 \\ \mathrm{Index}\mathrm{form}=4x^4+\mathrm{x}{\mathrm{ten}}^{\mathrm{three}}-3{\mathrm{10}}^2+2\mathrm{10}-21 \\ \mathrm{Coefficient}\mathrm{course}=\left(\mathrm{four},10,-3,\mathrm{two},-21\right) \\ \mathrm{Comparing}\mathrm{the}\mathrm{divisor}\mathrm{10}+\mathrm{iii}\mathrm{with}\mathrm{ten}-a,\mathrm{we}\mathrm{get}a=-3 \end{gathered}$$

$$\begin{gathered} \therefore \mathrm{Quotient}\mathrm{in}\mathrm{coefficient}\mathrm{form}=\left(4,-\mathrm{ii},\mathrm{iii},-\mathrm{vii}\right) \\ \mathrm{i}.\mathrm{e}.,\mathrm{caliber}=4{\mathrm{10}}^{\mathrm{iii}}-2x^2+3x-7\mathrm{in}\mathrm{the}\mathrm{variable}\mathrm{ten} \\ \mathrm{Here},\mathrm{the}\mathrm{remainder}\mathrm{is}0\left(\mathrm{aught}\right). \\ \therefore \left(x+\mathrm{three}\right)\mathrm{is}\mathrm{a}\mathrm{factor}\mathrm{of}\left(4{\mathrm{10}}^4+\mathrm{ten}{x}^3-\mathrm{iii}{x}^{\mathrm{two}}+2\mathrm{ten}-21\right). \end{gathered}$$

Folio No 173:

Question 11.iv:

(iiim ⁱⁱ + m $-$ 10) $÷$ (yard + 2)

Answer:

$$\begin{gathered} \mathrm{Dividend}=3m^2+g-10 \\ \mathrm{Index}\mathrm{grade}=\mathrm{three}{\mathrm{grand}}^{\mathrm{ii}}+m-\mathrm{x} \\ \mathrm{Coefficient}\mathrm{form}=\left(\mathrm{three},1,-10\right) \\ \mathrm{Comparing}\mathrm{the}\mathrm{divisor}\mathrm{chiliad}+2\mathrm{with}m-a,\mathrm{we}\mathrm{get}a=-2 \end{gathered}$$

$$\begin{gathered} \therefore \mathrm{Quotient}\mathrm{in}\mathrm{coefficient}\mathrm{grade}=\left(\mathrm{three},-5\right) \\ \mathrm{i}.\mathrm{eastward}.,\mathrm{quotient}=3g-\mathrm{five}\mathrm{in}\mathrm{the}\mathrm{variable}m \\ \mathrm{Remainder}=0 \\ \mathrm{Here},\mathrm{the}\mathrm{remainder}\mathrm{is}0\left(\mathrm{zippo}\right). \\ \therefore \left(m+2\right)\mathrm{is}\mathrm{a}\mathrm{factor}\mathrm{of}\left(3k^{\mathrm{ii}}+m-10\right). \end{gathered}$$

Page No 174:

Question 12:

Observe the zero in each of the polynomial given below:
(i) p(ten) = 9x $-$ 3
(ii) p(y) = 5y + 25

Respond:

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{To}\mathrm{find}\mathrm{the}\mathrm{zero}\mathrm{of}p\left(x\right),\mathrm{we}\mathrm{will}\mathrm{solve}\mathrm{the}\mathrm{equation}p\left(\mathrm{ten}\right)=0. \\ \mathrm{We}\mathrm{have}\mathrm{nine}\mathrm{10}-\mathrm{three}=0 \\ \Rightarrow 9x=3 \\ \therefore x=\frac{\mathrm{three}}{9}=\frac{1}{3} \\ \therefore \mathrm{The}\mathrm{zero}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{polynomial}\mathrm{is}\frac{1}{3}. \end{gathered}$$ $$\begin{gathered} \left(\mathrm{two}\right)\mathrm{To}\mathrm{find}\mathrm{the}\mathrm{goose\; egg}\mathrm{of}p\left(y\right),\mathrm{we}\mathrm{will}\mathrm{solve}\mathrm{the}\mathrm{equation}p\left(y\right)=0. \\ \mathrm{We}\mathrm{have}5y+25=0 \\ \Rightarrow 5y=-25 \\ \therefore y=\frac{-25}{\mathrm{five}}=-\mathrm{v} \\ \therefore \mathrm{The}\mathrm{zero}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{polynomial}\mathrm{is}-\mathrm{five}. \end{gathered}$$

Page No 174:

Question thirteen:

Find the value of the polynomial 2a ² $-$ 5a ³ + 7a $-$ 3 for a = 0, 2 and $-$ 1

Answer:

Let p(a) = 2a ²

$-$

5a ^three + sevena

$-$

iii

$$\begin{gathered} \mathrm{For}\mathrm{finding}\mathrm{the}\mathrm{value}\mathrm{of}p\left(a\right)\mathrm{when}a=0,\mathrm{put}a=0\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{polynomial}. \\ \mathrm{i}.\mathrm{eastward}.,p\left(0\right)=\mathrm{ii}{\left(0\right)}^2-\mathrm{five}{\left(0\right)}^3+7\left(0\right)-3 \\ =-3 \\ \therefore \mathrm{When}a=0,\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{polynomial}\mathrm{is}-3. \end{gathered}$$ $$\begin{gathered} \mathrm{Similarly},\mathrm{when}a=\mathrm{two},\mathrm{we}\mathrm{accept}: \\ p\left(\mathrm{ii}\right)=2{\left(2\right)}^{\mathrm{two}}-5{\left(2\right)}^3+7\left(\mathrm{ii}\right)-3 \\ =2\times 4-\mathrm{v}\times 8+14-3 \\ \Rightarrow p\left(2\right)=-21 \\ \therefore \mathrm{When}a=\mathrm{two},\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{polynomial}\mathrm{is}-21. \end{gathered}$$ $$\begin{gathered} \mathrm{Again},\mathrm{when}a=-1,\mathrm{nosotros}\mathrm{have}: \\ p\left(-1\right)=\mathrm{ii}{\left(-1\right)}^2-\mathrm{five}{\left(-\mathrm{one}\right)}^{\mathrm{iii}}+\mathrm{seven}\left(-1\right)-\mathrm{three} \\ =2+5-7-3 \\ \Rightarrow p\left(-1\right)=-3 \\ \therefore \mathrm{When}a=-1,\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{polynomial}\mathrm{is}-3. \end{gathered}$$

Page No 174:

Question 14:

If the value of the polynomial x ³ + twox ⁱⁱ $-$ ax + 1 at ten = ii is 11, find a.

Respond:

$$\begin{gathered} \mathrm{Let}p\left(x\right)={\mathrm{10}}^3+2{\mathrm{10}}^2-ax+1 \\ \mathrm{Then}p\left(2\right)=2^{\mathrm{three}}+2{\left(2\right)}^{\mathrm{two}}-a\left(\mathrm{two}\right)+1 \\ =\mathrm{eight}+\mathrm{viii}-2a+1 \\ \therefore p\left(2\right)=-2a+17 \\ \mathrm{Only}p\left(\mathrm{ii}\right)=11\left(\mathrm{Given}\right) \\ \mathrm{Now},-\mathrm{two}a+17=\mathrm{xi} \\ \Rightarrow -2a=11-17 \\ \Rightarrow -2a=-6 \\ \therefore a=\mathrm{iii} \end{gathered}$$

Page No 174:

Question 15:

If a $-$ 2y  + 5y ² is divided by (y $-$ 2) the remainder is 7, then notice the value of a.

Answer:

$$\begin{gathered} \mathrm{Let}p\left(y\right)=a-2y+5y^2 \\ \mathrm{Here},\mathrm{divisor}=y-\mathrm{ii} \\ \mathrm{When}y-2=0,\mathrm{we}\mathrm{take}y=\mathrm{two} \\ \mathrm{Putting}y=2\mathrm{in}p\left(y\right),\mathrm{we}\mathrm{get}: \\ \mathrm{Remainder}=p\left(2\right)=a-2\left(2\right)+5{\left(\mathrm{two}\right)}^{\mathrm{two}} \\ =a-4+20 \\ =a+\mathrm{sixteen} \\ \mathrm{Given}:\mathrm{Remainder}=7 \\ \mathrm{Now},a+16=7 \\ \Rightarrow a=7-\mathrm{xvi} \\ \therefore a=-9 \end{gathered}$$

Page No 174:

Question 16:

When 2ten ² $-$ ax + 7 and ax ² + 7x + 12 are divided by (10 $-$ 3) and (ten + ane) respectively, the remainder is aforementioned. Observe a.

Answer:

$$\begin{gathered} \mathrm{Let}p\left(x\right)=2{\mathrm{ten}}^2-ax+7 \\ \mathrm{Here},\mathrm{divisor}=x-\mathrm{three} \\ \mathrm{When}x-\mathrm{three}=0,\mathrm{we}\mathrm{have}x=\mathrm{iii} \\ \mathrm{Putting}x=3\mathrm{in}p\left(x\right),\mathrm{we}\mathrm{go}: \\ \mathrm{Remainder}=p\left(\mathrm{three}\right)=2{\left(\mathrm{three}\right)}^2-a\left(\mathrm{iii}\right)+7 \\ =\mathrm{eighteen}-3a+7 \\ =-\mathrm{three}a+25 \\ \mathrm{Similarly},\mathrm{let}q\left(x\right)=a{\mathrm{ten}}^2+7\mathrm{10}+12 \\ \mathrm{Here},\mathrm{divisor}=\mathrm{ten}+\mathrm{one} \\ \mathrm{When}x+1=0,\mathrm{we}\mathrm{have}x=-\mathrm{i} \\ \mathrm{Putting}x=-\mathrm{one}\mathrm{in}q\left(x\right),\mathrm{we}\mathrm{go}: \\ \mathrm{Remainder}=q\left(-\mathrm{ane}\right)=a{\left(-\mathrm{ane}\right)}^2+7\left(-1\right)+12 \\ =a-7+12 \\ =a+\mathrm{five} \\ \mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}\mathrm{the}\mathrm{remainders}\mathrm{in}\mathrm{both}\mathrm{the}\mathrm{cases}\mathrm{are}\mathrm{the}\mathrm{aforementioned}. \\ \mathrm{i}.\mathrm{due\; east}.,-3a+25=a+\mathrm{five} \\ \Rightarrow -3a-a=5-25 \\ \Rightarrow -4a=-20 \\ \therefore a=\mathrm{v} \end{gathered}$$

Page No 174:

Question 17:

If 2x + 1 is a factor of (3b + 2)x³ + (b $-$ i) then find b.

Answer:

$$\begin{gathered} \mathrm{Let}p\left(x\right)=\left(3b+2\right)x^3+\left(b-\mathrm{ane}\right) \\ \mathrm{Given}:2\mathrm{10}+1\mathrm{is}\mathrm{a}\mathrm{factor}\mathrm{of}p\left(x\right). \\ \therefore \mathrm{When}\mathrm{nosotros}\mathrm{divide}p\left(x\right)\mathrm{past}\mathrm{ii}\mathrm{ten}+1,\mathrm{the}\mathrm{residue}\mathrm{will}\mathrm{be}\mathrm{zero}. \\ \mathrm{Here},\mathrm{divisor}=2x+\mathrm{one} \\ \mathrm{When}2x+\mathrm{i}=0,\mathrm{we}\mathrm{have}\mathrm{10}=-\frac{1}{\mathrm{ii}} \\ \therefore \mathrm{Residuum}=p\left(-\frac{\mathrm{one}}{2}\right)=0 \\ \Rightarrow \left(\mathrm{iii}b+2\right){\left(-\frac{1}{2}\right)}^{\mathrm{iii}}+\left(b-\mathrm{i}\right)=0 \\ \Rightarrow \left(3b+2\right)\left(-\frac{\mathrm{ane}}{\mathrm{eight}}\right)+\left(b-1\right)=0 \\ \Rightarrow \frac{-\left(\mathrm{three}b+2\right)+\mathrm{eight}\left(b-\mathrm{ane}\right)}{8}=0 \\ \Rightarrow -\left(3b+\mathrm{two}\right)+\mathrm{eight}\left(b-1\right)=0 \\ \Rightarrow -\mathrm{iii}b-\mathrm{two}+8b-8=0 \\ \Rightarrow \mathrm{five}b-10=0 \\ \Rightarrow 5b=\mathrm{x} \\ \therefore b=2 \end{gathered}$$

Folio No 174:

Question xviii:

R_ane and R_ii are the remainders when the polynomial ax ³ + 3x _² $-$ iii and iix _³ $-$ fivex + twoa are divided by (x $-$ 4) respectively. If 2R₁ $-$ R₂ = 0, then find the value of a.

Answer:

$$\begin{gathered} \mathrm{Let}p\left(x\right)=a{x}^3+3{\mathrm{ten}}^{\mathrm{two}}-\mathrm{three} \\ \mathrm{Hither},\mathrm{divisor}=\mathrm{10}-4 \\ \mathrm{When}x-\mathrm{four}=0,\mathrm{we}\mathrm{take}x=4 \\ \mathrm{Given}:\mathrm{When}p\left(x\right)\mathrm{is}\mathrm{divided}\mathrm{by}\left(\mathrm{ten}-\mathrm{four}\right),\mathrm{the}\mathrm{remainder}\mathrm{is}{R}_{1.} \\ \therefore \mathrm{Remainder}=R_1=p\left(4\right) \\ \Rightarrow R_1=a{\left(4\right)}^{\mathrm{three}}+3{\left(4\right)}^2-\mathrm{iii} \\ \Rightarrow R_1=64a+48-3 \\ \Rightarrow R_{\mathrm{one}}=64a+45...\left(1\right) \\ \mathrm{Again},\mathrm{let}q\left(x\right)=2x^3-5\mathrm{10}+2a \\ \mathrm{Hither},\mathrm{divisor}=x-4 \\ \mathrm{When}\mathrm{10}-\mathrm{iv}=0,\mathrm{we}\mathrm{have}\mathrm{10}=\mathrm{four} \\ \mathrm{Given}:\mathrm{When}q\left(\mathrm{ten}\right)\mathrm{is}\mathrm{divided}\mathrm{past}\left(\mathrm{ten}-\mathrm{four}\right),\mathrm{the}\mathrm{rest}\mathrm{is}{R}_{\mathrm{ii}.} \\ \therefore \mathrm{Balance}=R_2=q\left(\mathrm{four}\right) \\ \Rightarrow R_2=2{\left(4\right)}^3-\mathrm{five}\left(4\right)+\mathrm{two}a \\ \Rightarrow R_2=128-20+2a \\ \Rightarrow R_{\mathrm{ii}}=\mathrm{ii}a+108...\left(\mathrm{two}\right) \\ \mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}2R_1-R_2=0 \\ \mathrm{Putting}\mathrm{the}\mathrm{values}\mathrm{of}{R}_1\mathrm{and}{R}_2\mathrm{from}\left(1\right)\mathrm{and}\left(\mathrm{two}\right),\mathrm{we}\mathrm{get}: \\ 2\left(64a+45\right)-\left(\mathrm{ii}a+108\right)=0 \\ \Rightarrow 128a+90-2a-108=0 \\ \Rightarrow 126a-\mathrm{eighteen}=0 \\ \Rightarrow 126a=18 \\ \Rightarrow a=\frac{18}{126}=\frac{1}{\mathrm{seven}} \\ \therefore a=\frac{\mathrm{i}}{7} \end{gathered}$$

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