Exercise 12.3 (Revised) - Chapter 14 - Factorisation - Ncert Solutions class 8 - Maths

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Chapter 12: Factorisation - NCERT Solutions for Class 8 Maths

Ex 12.3 Question 1.

Carry out the following divisions:
(i) $28 x^4 \div 56 x$
(ii) $-36 y^3 \div 9 y^2$
(iii) $66 p q^2 r^3 \div 11 q r^2$
(iv) $34 x^3 y^3 z^3 \div 51 x y^2 z^3$
(v) $12 a^8 b^8 \div\left(-6 a^6 b^4\right)$

Answer.

(i) $28 x^4 \div 56 x=\frac{28 x^4}{56 x}$
$
\begin{aligned}
& =\frac{28}{56} \times \frac{x^4}{x} \\
& =\frac{1}{2} x^3 \quad\left[\because x^m \div x^n=x^{m-n}\right]
\end{aligned}
$
(ii) $-36 y^3 \div 9 y^2=\frac{-36 y^3}{9 y^2}$
$
=\frac{-36}{9} \times \frac{y^3}{y^2}
$

$
=-4 y \quad\left[\because x^m \div x^n=x^{m-n}\right]
$
(iii) $66 p q^2 r^3 \div 11 q r^2$

$
\begin{aligned}
& =\frac{66 p q^2 r^3}{11 q r^2} \\
& =\frac{66}{11} \times \frac{p q^2 r^3}{q r^2} \\
& =6 p q r \quad\left[\because x^n \div x^n=x^{m-n}\right]
\end{aligned}
$
(iv) $34 x^3 y^3 z^3 \div 51 x y^2 z^3$
$
\begin{aligned}
& =\frac{34 x^3 y^3 z^3}{51 x y^2 z^3} \\
& =\frac{34}{51} \times \frac{x^3 y^3 z^3}{x y^2 z^3} \\
& =\frac{2}{3} x^2 y \quad\left[\because x^n \div x^n=x^{m-n}\right]
\end{aligned}
$
(v) $12 a^8 b^8 \div\left(-6 a^6 b^4\right)$
$
=\frac{12 a^8 b^8}{-6 a^5 b^4}
$

$
\begin{aligned}
= & \frac{12}{-6} \times \frac{a^8 b^8}{a^6 b^4} \\
= & -2 a^2 b^4 \quad\left[\because x^m \div x^n=x^{m-n}\right]
\end{aligned}
$
Ex 12.3 Question 2.

Divide the given polynomial by the given monomial:
(i) $\left(5 x^2-6 x\right) \div 3 x$
(ii) $\left(3 y^8-4 y^6+5 y^4\right) \div y^4$

(iii) $8\left(x^3 y^2 z^2+x^2 y^3 z^2+x^2 y^2 z^3\right) \div 4 x^2 y^2 z^2$
(iv) $\left(x^3+2 x^2+3 x\right) \div 2 x$
(v) $\left(p^3 q^6-p^6 q^3\right) \div p^3 q^3$

Answer.

(i) $\left(5 x^2-6 x\right) \div 3 x$
$
\begin{aligned}
& =\frac{5 x^2-6 x}{3 x} \\
& =\frac{5 x^2}{3 x}-\frac{6 x}{3 x}=\frac{5}{3} x-2=\frac{1}{3}(5 x-6)
\end{aligned}
$
$
\text { (ii) } \begin{aligned}
& \left(3 y^8-4 y^6+5 y^4\right) \div y^4 \\
= & \frac{3 y^8-4 y^6+5 y^4}{y^4} \\
= & \frac{3 y^8}{y^4}-\frac{4 y^6}{y^4}+\frac{5 y^4}{y^4}=3 y^4-4 y^2+5
\end{aligned}
$
(iii) $8\left(x^3 y^2 z^2+x^2 y^3 z^2+x^2 y^2 z^3\right) \div 4 x^2 y^2 z^2$

$
\begin{aligned}
= & \frac{8\left(x^3 y^2 z^2+x^2 y^3 z^2+x^2 y^2 z^3\right)}{4 x^2 y^2 z^2} \\
& =\frac{8 x^3 y^2 z^2}{4 x^2 y^2 z^2}+\frac{8 x^2 y^3 z^2}{4 x^2 y^2 z^2}+\frac{8 x^2 y^2 z^3}{4 x^2 y^2 z^2} \\
& =2 x+2 y+2 z \\
& =2(x+y+z)
\end{aligned}
$
(iv) $\left(x^3+2 x^2+3 x\right) \div 2 x$

$
\begin{aligned}
& =\frac{x^3+2 x^2+3 x}{2 x} \\
& =\frac{x^3}{2 x}+\frac{2 x^2}{2 x}+\frac{3 x}{2 x}=\frac{x^2}{2}+\frac{2 x}{2}+\frac{3}{2} \\
& =\frac{1}{2}\left(x^2+2 x+3\right)
\end{aligned}
$
$
\text { (v) } \begin{aligned}
& \left(p^3 q^6-p^6 q^3\right) \div p^3 q^3 \\
& =\frac{p^3 q^6-p^6 q^3}{p^3 q^3} \\
& =\frac{p^3 q^6}{p^3 q^3}-\frac{p^6 q^3}{p^3 q^3}=q^3-p^3
\end{aligned}
$
Ex 12.3 Question 3.

Work out the following divisions:
(i) $(10 x-25) \div 5$
(ii) $(10 x-25) \div(2 x-5)$
(iii) $10 y(6 y+21) \div 5(2 y+7)$

(iv) $9 x^2 y^2(3 z-24) \div 27 x y(z-8)$
(v) $96 a b c(3 a-12)(5 b-30) \div 144(a-4)(b-6)$

Answer.

(i) $(10 x-25) \div 5=\frac{10 x-25}{5}$
$
=\frac{5(2 x-5)}{5}=2 x-5
$

(ii) $(10 x-25) \div(2 x-5)=\frac{10 x-25}{(2 x-5)}$
$
=\frac{5(2 x-5)}{(2 x-5)}=5
$
(iii) $10 y(6 y+21) \div 5(2 y+7)$
$
\begin{aligned}
& =\frac{10 y(6 y+21)}{5(2 y+7)} \\
& =\frac{2 \times 5 \times y \times 3(2 y+7)}{5(2 y+7)}=2 \times y \times 3=6 y
\end{aligned}
$
(iv) $9 x^2 y^2(3 z-24) \div 27 x y(z-8)$
$
\begin{aligned}
= & \frac{9 x^2 y^2(3 z-24)}{27 x y(z-8)} \\
= & \frac{9}{27} \times \frac{x y \times x y \times 3(z-8)}{x y(z-8)}=x y
\end{aligned}
$
(v) $96 a b c(3 a-12)(5 b-30) \div 144(a-4)(b-6)$

$
\begin{aligned}
& =\frac{96 a b c(3 a-12)(5 b-30)}{144(a-4)(b-6)} \\
& =\frac{12 \times 4 \times 2 \times a b c \times 3(a-4) \times 5(b-6)}{12 \times 4 \times 3(a-4)(b-6)} \\
& =10 a b c
\end{aligned}
$
Ex 12.3 Question 4.

Divide as directed:

(i) $5(2 x+1)(3 x+5) \div(2 x+1)$
(ii) $26 x y(x+5)(y-4) \div 13 x(y-4)$
(iii) $52 p q r(p+q)(q+r)(r+p) \div 104 p q(q+r)(r+p)$
(iv) $20(y+4)\left(y^2+5 y+3\right) \div 5(y+4)$
(v) $x(x+1)(x+2)(x+3) \div x(x+1)$

Answer.

(i) $5(2 x+1)(3 x+5) \div(2 x+1)$
$
\begin{aligned}
& =\frac{5(2 x+1)(3 x+5)}{(2 x+1)} \\
& =5(3 x+5)
\end{aligned}
$
$
\begin{aligned}
& \text { (ii) } 26 x y(x+5)(y-4) \div 13 x(y-4) \\
& 26 x y(x+5)(y-4) \div 13 x(y-4) \\
& =\frac{26 x y(x+5)(y-4)}{13 x(y-4)}
\end{aligned}
$

$
=\frac{13 \times 2 \times x y(x+5)(y-4)}{13 x(y-4)}=2 y(x+5)
$
(iii)
$
\begin{aligned}
& 52 p q r(p+q)(q+r)(r+p) \div 104 p q(q+r)(r+p) \\
& =\frac{52 p q r(p+q)(q+r)(r+p)}{52 \times 2 \times p q(q+r)(r+p)} \\
& =\frac{1}{2} r(p+q)
\end{aligned}
$

(iv)
$
\begin{aligned}
& 20(y+4)\left(y^2+5 y+3\right) \div 5(y+4) \\
& =\frac{20(y+4)\left(y^2+5 y+3\right)}{5(y+4)} \\
& =4\left(y^2+5 y+3\right)
\end{aligned}
$
(v)
$
\begin{aligned}
& x(x+1)(x+2)(x+3) \div x(x+1) \\
& =\frac{x(x+1)(x+2)(x+3)}{x(x+1)} \\
& =(x+2)(x+3)
\end{aligned}
$
Ex 12.3 Question 5.

Factorize the expressions and divide them as directed:
(i) $\left(y^2+7 y+10\right) \div(y+5)$

(ii) $\left(m^2-14 m-32\right) \div(m+2)$
(iii) $\left(5 p^2-25 p+20\right) \div(p-1)$
(iv) $4 y z\left(z^2+6 z-16\right) \div 2 y(z+8)$
(v) $5 p q\left(p^2-q^2\right) \div 2 p(p+q)$
(vi) $12 x y\left(9 x^2-16 y^2\right) \div 4 x y(3 x+4 y)$
(vii) $39 y^3\left(50 y^2-98\right) \div 26 y^2(5 y+7)$

Answer.

(i) $\left(y^2+7 y+10\right) \div(y+5)$
$
\begin{aligned}
& =\frac{y^2+7 y+10}{(y+5)} \\
& =\frac{y^2+(2+5) y+2 \times 5}{(y+5)} \\
& =\frac{y^2+2 y+5 y+2 \times 5}{(y+5)} \\
& =\frac{(y+2)(y+5)}{(y+5)}\left[\because x^2+(a+b) x+a b=(x+a)(x+b)\right] \\
& =y+2
\end{aligned}
$
$
\begin{aligned}
& \text { (ii) }\left(m^2-14 m+32\right) \div(m+2) \\
& =\frac{m^2-14 m+32}{(m+2)} \\
& =\frac{m^2+(-16+2) m+(-16) \times 2}{(m+2)}
\end{aligned}
$

$\begin{aligned}
&\begin{aligned}
& =\frac{(m-16)(m+2)}{(m+2)}\left[\because x^2+(a+b) x+a b=(x+a)(x+b)\right] \\
& =(m-16)
\end{aligned}\\
&\begin{aligned}
& \text { (iii) }\left(5 p^2-25 p+20\right) \div(p-1) \\
& =\frac{5 p^2-25 p+20}{(p-1)}
\end{aligned}
\end{aligned}$

$\begin{aligned}
&\begin{aligned}
& =\frac{5 p^2-20 p-5 p+20}{(p-1)} \\
& =\frac{5 p(p-4)-5(p-4)}{(p-1)} \\
& =\frac{(5 p-5)(p-4)}{(p-1)}=\frac{5(p-1)(p-4)}{(p-1)} \\
& =5(p-4)
\end{aligned}\\
&\begin{aligned}
& \text { (iv) } 4 y z\left(z^2+6 z-16\right) \div 2 y(z+8) \\
& =\frac{4 y z\left(z^2+6 z-16\right)}{2 y(z+8)} \\
& =\frac{4 y z\left[z^2+(8-2) z+8 \times(-2)\right]}{2 y(z+8)} \\
& =\frac{4 y z(z-2)(z+8)}{2 y(z+8)}\left[\because x^2+(a+b) x+a b=(x+a)(x+b)\right] \\
& =2 z(z-2)
\end{aligned}
\end{aligned}$

$\text { (v) } \begin{aligned}
& 5 p q\left(p^2-q^2\right) \div 2 p(p+q) \\
& =\frac{5 p q\left(p^2-q^2\right)}{2 p(p+q)} \\
& =\frac{5 p q(p-q)(p+q)}{2 p(p+q)}\left[\because a^2-b^2=(a-b)(a+b)\right] \\
& =\frac{5}{2} q(p-q)
\end{aligned}$

$
\text { (vi) } \begin{aligned}
& 12 x y\left(9 x^2-16 y^2\right)-4 x y(3 x+4 y) \\
& =\frac{12 x y\left(9 x^2-16 y^2\right)}{4 x y(3 x+4 y)} \\
& =\frac{12 x y\left[(3 x)^2-(4 y)^2\right]}{4 x y(3 x+4 y)} \\
& =\frac{12 x y(3 x-4 y)(3 x+4 y)}{4 x y(3 x+4 y)}\left[\because a^2-b^2=(a-b)(a+b)\right] \\
& =3(3 x-4 y)
\end{aligned}
$
(vii) $39 y^3\left(50 y^2-98\right) \div 26 y^2(5 y+7)$
$
\begin{aligned}
= & \frac{39 y^3\left(50 y^2-98\right)}{26 y^2(5 y+7)} \\
= & \frac{39 y^3 \times 2\left(25 y^2-49\right)}{26 y^2(5 y+7)}
\end{aligned}
$

$\begin{aligned}
& =\frac{39 y^2 \times 2\left[(5 y)^2-(7)^2\right]}{26 y^2(5 y+7)} \text { change the image with image_3312_1 } \\
& =\frac{39 y^2 \times 2(5 y-7)(5 y+7)}{26 y^2(5 y+7)}\left[\because a^2-b^2=(a-b)(a+b)\right] \text { change the image with } \\
& \text { image_3312_2 } \\
& =3 y(5 y-7)
\end{aligned}$