ML AGGARWAL CLASS 8 CHAPTER 11 Factorisation Exercise 11.1

 Exercise 11.1

Solution 1

Sol :

(i) $8 x y^{3}+12 x^{2} y^{2}$

H.C.F of $8 x y^{3}$ and $12 x^{2} y^{2}$ is $4 x y^{2}$

$\therefore$ Divide each expression by $4 x y^{2}$ and keep $4 x y^{2}$
outside the bracket.

$\Rightarrow \quad 4 x y^{2}(2 y+3 x)$

$\therefore 8 x y^{3}+12 x^{2} y^{2}=4 x y^{2}(2 y+3 x)$


(ii) $15 a x^{3}-9 a x^{2}$

H.C.F of expression $15 a x^{3}$ and $9 a x^{2}$ is $3 a x^{2}$

$\quad 15 a x^{3}-9 a x^{2}=3 a x^{2}(5 x-3)$

Solution 2

Sol :
(i) $21 p y^{2}-56 p y$

$H \cdot C \cdot F$ of $21 p y^{2}$ and $56 p y$ is $7 p y$

$\therefore 21 p y^{2}-56 p y \Rightarrow 7 p y(3 y-8)$


(ii) $4 x^{3}-6 x^{2}$

H.C.F of $4 x^{3}$ and $6 x^{2}$ is $2 x^{2}$

$\therefore 4 x^{3}-6 x^{2} \Rightarrow 2 x^{2}(2 x-3)$

Solution 3

Sol :

(i) $25 a b c^{2}-15 a^{2} b^{2} c$

H.C.F of $25 a b c^{2}$ and $15 a^{2} b^{2} c$ is $5 a b c$

$\therefore 25 a b c^{2}-15 a^{2} b^{2} c \Rightarrow 5 a b c(5 c-3 a b)$


(ii)
 $x^{2} y z+x y^{2} z+x y z^{2}$

H.C.F of $x^{2} y z$, $x y^{2} z$ and $x y z^{2}$ is $x y z$

$\therefore \quad x y z(x+y+z)$

Solution 4

Sol :
(i) $8 x^{3}-6 x^{2}+10 x$

H.C.F of $8 x^{3}, 6 x^{2}$ and $10 x$ is $2 x$

$\Rightarrow \quad 2 x\left(4 x^{2}-3 x+5\right)$


(ii) $14 \mathrm{mn}+22 \mathrm{~m}-6 \mathrm{p}$

H.C.F of $14 \mathrm{mn}, 22 \mathrm{~m}$ and $62 \mathrm{P}$ is 2

$\Rightarrow 2(7 \mathrm{mn}+11 \mathrm{n}-31 \mathrm{P})$


Solution 5

Sol :
(i) $18 p^{2} q^{2}-24 p q^{2}+30 p^{2} q$

$H \cdot C \cdot F$ of $18 p^{2} q^{2}, 24 p a^{2}$ and $30 p^{2} q$ is $6 p q$

$\Rightarrow \quad 6 p q(3 p q-4 q+5 p)$


(ii) $27 a^{3} b^{3}-18 a^{2} b^{3}+75 a^{3} b^{2}$

H. C .F of $27 a^{3} b^{3}, 18 a^{2} b^{2}$ and $7 s a^{3} b^{2}$ is $3 a^{2} b^{2}$

$\Rightarrow 3 a^{2} b^{2}(9 a b-6 b+25 a)

Solution 6

Sol :
(i) 15a(2p-3q)-10b(2p-3q)

HCFof 15 a(2 p-3 q) and 10b(2 p-3 q) is 5(2 p-3 q)

$\Rightarrow 5(2 p-3 q)(3 a-2 b)$


(ii) $3 a\left(x^{2}+y^{2}\right)+6 b\left(x^{2}+y^{2}\right)$

HCF of $3 a\left(x^{2}+y^{2}\right)$ and $6 b\left(x^{2}+y^{2}\right)$ is $3\left(x^{2}+y^{2}\right)$

$\therefore \Rightarrow 3\left(x^{2}+y^{2}\right)(a+2 b)$

Solution 7

Sol :
(i) $6(x+2 y)^{3}+8(x+2 y)^{2}$

H.C.F of $6(x+2 y)^{3}$ and $8(x+2 y)^{2}$ is $2(x+2 y)^{2}$

$\therefore \quad 2(x+2 y)^{2} \quad(3(x+2 y)+4)$

$\Rightarrow \quad 2(x+2 y)^{2}(3 x+6 y+4)$



(ii)$14(a-3 b)^{3}-21 p(a-3 b)$

H.C.F of $14(a-3 b)^{3}$ and 21p(a-3 b) is F(a-3 b)

$\rightarrow \quad 7(a-3 b)\left[2(a-3 b)^{2}-3 p\right]$

Solution 8

Sol :
(i) $\quad 10 a(2 p+q)^{3}-15 b(2 p+q)^{2}+35(2 p+q)$

HCF of $10 a(2 p+q)^{3}, 15 b(2 p+q)^{2}$ and

35(2p+q) is 5(2p+q)

$\Rightarrow 5(2 p+q)(2 a-3 b+7)$

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