Exercise 12.2 (Revised) - Chapter 14 - Factorisation - Ncert Solutions class 8 - Maths

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Chapter 12: Factorisation - NCERT Solutions for Class 8 Maths

Ex 12.2 Question 1.

Factorize the following expressions:
(i) $a^2+8 a+16$
(ii) $p^2-10 p+25$
(iii) $25 m^2+30 m+9$
(iv) $49 y^2+84 y z+36 z^2$
(v) $4 x^2-8 x+4$
(vi) $121 b^2-88 b c+16 c^2$
(vii) $(l+m)^2-4 l m$
[Hint: Expand $(I+m)^2$ first]
(viii) $a^4+2 a^2 b^2+b^4$

Answer.

(i) $a^2+8 a+16=a^2+(4+4) a+4 \times 4$
Using identity $x^2+(a+b) x+a b=(x+a)(x+b)$,
Here $x=a: a=4$ and $b=4$
$
a^2+8 a+16=(a+4)(a+4)=(a+4)^2
$

(ii) $p^2-10 p+25=p^2+(-5-5) p+(-5)(-5)$

Using identity $x^2+(a+b) x+a b=(x+a)(x+b)$,

Here $x=p, a=-5$ and $b=-5$
$
p^2-10 p+25=(p-5)(p-5)=(p-5)^2
$
(iii) $25 m^2+30 m+9=(5 m)^2+2 \times 5 m \times 3+(3)^2$

Using identity $a^2+2 a b+b^2=(a+b)^2$, here $a=5 m, b=3$
$
25 m^2+30 m+9=(5 m+3)^2
$
(iv) $49 y^2+84 y z+36 z^2=(7 y)^2+2 \times 7 y \times 6 z+(6 z)^2$

Using identity $a^2+2 a b+b^2=(a+b)^2$, here $a=7 y, b=6 z$
$
49 y^2+84 y z+36 z^2=(7 y+6 z)^2
$
(v) $4 x^2-8 x+4=(2 x)^2-2 \times 2 x \times 2+(2)^2$

Using identity $a^2-2 a b+b^2=(a-b)^2$, here $a=2 x, b=2$
$
\begin{aligned}
& 4 x^2-8 x+4=(2 x-2)^2 \\
& =(2)^2(x-1)^2=4(x-1)^2
\end{aligned}
$

(vi) $121 b^2-88 b c+16 c^2=(11 b)^2-2 \times 11 b \times 4 c+(4 c)^2$

Using identity $a^2-2 a b+b^2=(a-b)^2$, here $a=11 b, b=4 c$
$
121 b^2-88 b c+16 c^2=(11 b-4 c)^2
$
(vii) $(l+m)^2-4 l m$
$
\begin{aligned}
= & l^2+2 \times l \times m+m^2-4 l m\left[\because(a+b)^2=a^2+2 a b+b^2\right] \\
& =l^2+2 l m+m^2-4 l m
\end{aligned}
$

$\begin{aligned}
&\begin{aligned}
& =l^2-2 l m+m^2 \\
& =(l-m)^2\left[\because \quad(a-b)^2=a^2-2 a b+b^2\right]
\end{aligned}\\
&\text { (viii) } \begin{aligned}
& a^4+2 a^2 b^2+b^4=\left(a^2\right)^2+2 \times a^2 \times b^2+\left(b^2\right)^2 \\
& =\left(a^2+b^2\right)^2 \quad\left[\because(a+b)^2=a^2+2 a b+b^2\right]
\end{aligned}
\end{aligned}$

Ex 12.2 Question 2.

Factorize:
(i) $4 p^2-9 q^2$
(ii) $63 a^2-112 b^2$
(iii) $49 x^2-36$
(iv) $16 x^5-144 x^2$
(v) $(l+m)^2-(l-m)^2$
(vi) $9 x^2 y^2-16$
(vii) $\left(x^2-2 x y+y^2\right)-z^2$
(viii) $25 a^2-4 b^2+28 b c-49 c^2$

Answer.

(i) $4 p^2-9 q^2=(2 p)^2-(3 q)^2$
$=(2 p-3 q)(2 p+3 q) \quad\left[\because a^2-b^2=(a-b)(a+b)\right]$
(ii) $63 a^2-112 b^2=7\left(9 a^2-16 b^2\right)$
$
=7\left[(3 a)^2-(4 b)^2\right]
$

$
=7(3 a-4 b)(3 a+4 b)\left[\because a^2-b^2=(a-b)(a+b)\right]
$
(iii) $49 x^2-36=(7 x)^2-(6)^2$
$
=(7 x-6)(7 x+6) \quad\left[\because a^2-b^2=(a-b)(a+b)\right]
$
$
\begin{aligned}
& \text { (iv) } 16 x^5-144 x^3=16 x^3\left(x^2-9\right) \\
& =16 x^3\left[(x)^2-(3)^2\right] \\
& =16 x^3(x-3)(x+3) \quad\left[\because a^2-b^2=(a-b)(a+b)\right]
\end{aligned}
$

$\begin{aligned}
&\begin{aligned}
& \text { (v) }(l+m)^2-(l-m)^2 \\
& =[(l+m)+(l-m)][(l+m)-(l-m)]\left[\because a^2-b^2=(a-b)(a+b)\right] \\
& =(l+m+l-m)(l+m-l+m) \\
& =(2 l)(2 \mathrm{~m})=4 \mathrm{~lm}
\end{aligned}\\
&\begin{aligned}
& \text { (vi) } 9 x^2 y^2-16=(3 x y)^2-(4)^2 \\
& =(3 x y-4)(3 x y+4) \quad\left[\because a^2-b^2=(a-b)(a+b)\right] \\
&
\end{aligned}\\
&\text { (vii) } \begin{aligned}
\left(x^2-2 x y+y^2\right)-z^2=(x-y)^2-z^2\left[\because(a-b)^2=a^2-2 a b+b^2\right] \\
=(x-y-z)(x-y+z)\left[\because a^2-b^2=(a-b)(a+b)\right]
\end{aligned}\\
&\text { (viii) } \begin{aligned}
& 25 a^2-4 b^2+28 b c-49 c^2 \\
= & 25 a^2-\left(4 b^2-28 b c+49 c^2\right) \\
= & 25 a^2-\left[(2 b)^2-2 \times 2 b \times 7 c+(7 c)^2\right]
\end{aligned}
\end{aligned}$

$
\begin{aligned}
& =25 a^2-(2 b-7 c)^2\left[\because(a-b)^2=a^2-2 a b+b^2\right] \\
& =(5 a)^2-(2 b-7 c)^2 \\
& =[5 a-(2 b-7 c)][5 a+(2 b-7 c)]\left[\because a^2-b^2=(a-b)(a+b)\right] \\
& =(5 a-2 b+7 c)(5 a+2 b-7 c)
\end{aligned}
$
Ex 12.2 Question 3.

Factorize the expressions:
(i) $a x^2+b x$
(ii) $7 p^2+21 q^2$
(iii) $2 x^3+2 x y^2+2 x z^2$
(iv) $a m^2+b m^2+b n^2+a n^2$
(v) $(I m+l)+m+1$
(vi) $y(y+z)+9(y+z)$
(vii) $5 y^2-20 y-8 z+2 y z$
(viii) $10 a b+4 a+5 b+2$

(ix) $6 x y-4 y+6-9 x$

Answer.

(i) $a x^2+b x=x(a x+b)$
(ii) $7 p^2+21 q^2=7\left(p^2+3 q^2\right)$
(iii) $2 x^3+2 x y^2+2 x z^2=2 x\left(x^2+y^2+z^2\right)$
(iv) $a m^2+b m^2+b n^2+a n^2$

$
\begin{gathered}
=m^2(a+b)+n^2(a+b) \\
=(a+b)\left(m^2+n^2\right)
\end{gathered}
$
(v) $(I m+l)+m+1=l(m+1)+1(m+1)$
$
=(m+1)(l+1)
$
(vi) $y(y+z)+9(y+z)=(y+z)(y+9)$
$
\begin{aligned}
\text { (vii) } & 5 y^2-20 y-8 z+2 y z \\
= & 5 y^2-20 y+2 y z-8 z \\
= & 5 y(y-4)+2 z(y-4) \\
= & (y-4)(5 y+2 z)
\end{aligned}
$
$
\begin{aligned}
& \text { (viii) } 10 a b+4 a+5 b+2 \\
& =2 a(5 b+2)+1(5 b+2) \\
& =(5 b+2)(2 a+1)
\end{aligned}
$
(ix) $6 x y-4 y+6-9 x$

$
\begin{aligned}
& =6 x y-9 x-4 y+6 \\
& =3 x(2 y-3)-2(2 y-3) \\
& =(2 y-3)(3 x-2)
\end{aligned}
$
Ex 12.2 Question 4.

Factorize:
(i) $a^4-b^4$
(ii) $p^4-81$

(iii) $x^4-(y+z)^4$
(iv) $x^4-(x-z)^4$
(v) $a^4-2 a^2 b^2+b^4$
$
\begin{aligned}
& \text { Ans. (i) } a^4-b^4=\left(a^2\right)^2-\left(b^2\right)^2 \\
& =\left(a^2-b^2\right)\left(a^2+b^2\right) \quad\left[\because a^2-b^2-(a-b)(a+b)\right] \\
& =(a-b)(a+b)\left(a^2+b^2\right) \quad\left[\because a^2-b^2-(a-b)(a+b)\right]
\end{aligned}
$
$
\text { (ii) } \begin{aligned}
& p^4-81=\left(p^2\right)^2-(9)^2 \\
= & \left(p^2-9\right)\left(p^2+9\right) \quad\left[\because a^2-b^2-(a-b)(a+b)\right] \\
= & \left(p^2-3^2\right)\left(p^2+9\right) \\
= & (p-3)(p+3)\left(p^2+9\right) \quad\left[\because a^2-b^2-(a-b)(a+b)\right]
\end{aligned}
$
$
\begin{aligned}
& \text { (iii) } x^4-(y+z)^4=\left(x^2\right)^2-\left[(y+z)^2\right]^2 \\
& =\left[x^2-(y+z)^2\right]\left[x^2+(y+z)^2\right]\left[\because a^2-b^2-(a-b)(a+b)\right] \\
& =[x-(y+z)][x+(y+z)]\left[x^2+(y+z)^2\right]\left[\because a^2-b^2-(a-b)(a+b)\right] \\
&
\end{aligned}
$

$\begin{aligned}
& \text { (iv) } x^4-(x-z)^4=\left(x^2\right)^2-\left[(x-z)^2\right]^2 \\
& =\left[x^2-(x-z)^2\right]\left[x^2+(x-z)^2\right]\left[\because a^2-b^2-(a-b)(a+b)\right] \\
& =[\mathrm{x}-(\mathrm{x}-\mathrm{z})][\mathrm{x}+(\mathrm{x}-\mathrm{z})]\left[\mathrm{x}^2+(\mathrm{x}-\mathrm{z})^2\right]\left[\because a^2-b^2-(a-b)(a+b)\right]
\end{aligned}$

$
\begin{aligned}
& =[\mathrm{x}-\mathrm{x}+\mathrm{z}][\mathrm{x}+\mathrm{x}-\mathrm{z}]\left[\mathrm{x}^2+\mathrm{x}^2-2 \mathrm{xz}+\mathrm{z}^2\right]\left[\because(a-b)^2=a^2-2 a b+b^2\right] \\
& =\mathrm{z}(2 \mathrm{x}-\mathrm{z})\left(2 \mathrm{x}^2-2 \mathrm{xz}+\mathrm{z}^2\right)
\end{aligned}
$
(v) $a^4-2 a^2 b^2+b^4=\left(a^2\right)^2-2 a^2 b^2+\left(b^2\right)^2$
$
\begin{aligned}
& =\left(a^2-b^2\right)^2 \quad\left[\because(a-b)^2=a^2-2 a b+b^2\right] \\
& =[(a-b)(a+b)]^2 \quad\left[\because a^2-b^2-(a-b)(a+b)\right] \\
& =(a-b)^2(a+b)^2 \quad\left[\because(x y)^m=x^m-y^m\right]
\end{aligned}
$
Ex 12.2 Question 5.

Factorize the following expressions:
(i) $p^2+6 p+8$
(ii) $q^2-10 q+21$
(iii) $p^2+6 p-16$

Answer.
(i) $p^2+6 p+8=p^2+(4+2) p+4 \times 2$

$\begin{aligned}
&\begin{aligned}
& =p^2+4 p+2 p+4 \times 2 \\
& =p(p+4)+2(p+4) \\
& =(p+4)(p+2)
\end{aligned}\\
&\text { (ii) } \begin{aligned}
& q^2-10 q+21=q^2-(7+3) q+7 \times 3 \\
= & q^2-7 q-3 q+7 \times 3
\end{aligned}
\end{aligned}$

$\begin{aligned}
&\begin{aligned}
& =q(q-7)-3(q-7) \\
& =(q-7)(q-3)
\end{aligned}\\
&\text { (iii) } \begin{aligned}
& p^2+6 p-16=p^2+(8-2) p-8 \times 2 \\
= & p^2+8 p-2 p-8 \times 2 \\
= & p(p+8)-2(p+8) \\
= & (p+8)(p-2)
\end{aligned}
\end{aligned}$