Exercise 3.5 - Chapter 3 Algebra 9th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

$\mathbf{E x 3 . 5}$
Question 1.
Factorise the following expressions:
(i) $2 \mathrm{a}^{2}+4 \mathrm{a}^{2} \mathrm{~b}+8 \mathrm{a}^{2} \mathrm{c}$
(ii) $a b-a c-m b+m c$
Solution:
(i) $2 \mathrm{a}^{2}+4 \mathrm{a}^{2} \mathrm{~b}+8 \mathrm{a}^{2} \mathrm{c}=2 \mathrm{a}^{2}[1+2 \mathrm{~b}+4 \mathrm{c}]$
(ii) $a b-a c-m b+m c=a(b-c)-m(b-c)=(b-c)(a-m)$

Question 2.

Factorise the following:
(i) $x^{2}+4 x+4$
(ii) $3 \mathrm{a}^{2}-24 \mathrm{ab}+48 \mathrm{~b}^{2}$
(iii) $x^{5}-16 x$
(iv) $m^{2}+\frac{1}{m^{2}}-23$
(v) $6-216 x^{2}$
(vi) $a^{2}+\frac{1}{a^{2}}-18$
Solution:
(i) $x^{2}+4 x+4=(x+2)(x+2)=(x+2)^{2}$
$\because(\mathrm{a}+\mathrm{b})^{2}=\mathrm{a}^{2}+2 \mathrm{ab}+\mathrm{b}^{2}$
(ii) $3 \mathrm{a}^{2}-24 \mathrm{ab}+48 \mathrm{~b}^{2}=3\left[\mathrm{a}^{2}-8 \mathrm{ab}+16 \mathrm{~b}^{2}\right.$ ]
$=3[a-4 b]^{2}\left(\because(a-b)^{2}=a^{2}-2 a b+b^{2}\right)$
(iii) $x^{5}-16 x=x\left[x^{4}-16\right]=x\left[\left(x^{2}\right)^{2}-4^{2}\right]$
$=x\left(x^{2}+4\right)\left(x^{2}-4\right)=x\left(x^{2}+4\right)(x+2)(x-2)$

(iv)$\begin{aligned}
m^{2}+\frac{1}{m^{2}}-23 &=\left(m+\frac{1}{m}\right)^{2}-2-23=\left(m+\frac{1}{m}\right)^{2}-25 \\
&=\left(m+\frac{1}{m}\right)^{2}-5^{2}=\left(m+\frac{1}{m}-5\right)\left(m+\frac{1}{m}+5\right)
\end{aligned}$

(v)$6-216 x^{2}=6\left(1-(6 x)^{2}\right)=6(1+6 x)(1-6 x)$

(vi)$a^{2}+\frac{1}{a^{2}}-18=\left(a-\frac{1}{a}\right)^{2}+2-18=\left(a-\frac{1}{a}\right)^{2}-16=\left(a-\frac{1}{a}+4\right)\left(a-\frac{1}{a}-4\right)$
 

Question 3. Factorise the following:

(i) $4 x^{2}+9 y^{2}+25 z^{2}+12 x y+30 y z+20 x z$
(ii) $25 x^{2}+4 y^{2}+9 z^{2}-20 x y+12 y z-30 x z$
Solution:
(i) $4 x^{2}+9 y^{2}+25 z^{2}+12 x y+30 y z+20 x z$
$=(2 x)^{2}+(3 y)^{2}+(5 z)^{2}+2(2 x)(3 y)+2(3 y)+(5 z)+2 \times 3 y \times 5 z$
$=(2 x+3 y+5 z)^{2}$
$\because(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a$
(ii) $25 \mathrm{x}^{2}+4 \mathrm{y}^{2}+9 \mathrm{z}^{2}-20 \mathrm{xy}+12 \mathrm{yz}-30 \mathrm{xz}$ $=(5 x)^{2}+(-2 y)^{2}+(-3 z)^{2}+2(5 x)(-2 y)+2(-2 y)(-3 z)+2(-3 z)(5 x)$ $=(5 x-2 y-3 z)^{2}$

Question $4 .$
Factorise the following
(i) $8 x^{3}+125 y^{3}$
(ii) $27 \mathrm{x}^{3}-8 \mathrm{y}^{3}$
(iii) $a^{6}-64$
Solution:
(i) $8 x^{3}+125 y^{3}(2 x)^{3}+(5 y)^{3}$
$\therefore \mathrm{a}^{3}-\mathrm{b}^{3}=(\mathrm{a}+\mathrm{b})\left(\mathrm{a}^{2}-\mathrm{ab}+\mathrm{b}^{2}\right)$
$=(2 x+5 y)\left[(2 x)^{2}-(2 x)(5 y)+\left(5 y^{2}\right]\right.$
$=(2 x+5 y)^{2}\left(4 x^{2}-10 x y+25 y^{2}\right)$
(ii) $27 \mathrm{x}^{3}-8 \mathrm{y}^{3}=(3 \mathrm{x})^{3}-(2 \mathrm{y})^{2}$
$=(3 x-2 y)\left((3 x)^{2}+3 x \times 2 y+(2 y)^{3}\right)$
$=(3 x-2 y)\left(9 x^{3}+6 x y+4 x y+4 y^{3}\right)$

(iii) $a^{6}-64=\left(a^{2}\right)^{3}-4^{3}\left(a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)\right.$ $=\left(a^{2}-4\right)\left(a^{4}+4 a^{2}+4^{2}\right)$ $=(a+2)(a-2)\left(a^{2}+4-2 a\right)\left(a^{2}-4+2 a\right)$ Question 5 . Factorise the following: (i) $x^{3}+8 y^{3}+6 x y-1$ (ii) $l^{3}-8 m^{3}-27 n^{3}-18 \operatorname{lmn}$ Solution: (i) $x^{3}+8 y^{3}+6 x y-1=x^{3}+(2 y)^{3}+(-1)^{3}-3(x)(2 y)(-1)$ $=(x+2 y-1)\left(x^{2}+4 y^{2}+1-2 x y+2 y+x\right)$ (ii) $l^{3}-8 m^{3}-27 n^{3}-18 \operatorname{lmn}=l^{3}+(-2 m)^{3}+(-3 n)^{3}-3(1)\{-2 m)(-3 n)$ $=(1-2 m-3 n)\left(l^{2}+(-2 m)^{2}+(-3 n)^{3}-1 \times-2 m-(-2 m \times-3 n)-(-3 n \times 1)\right)$ $=(1-2 m-3 n)\left(l^{2}+4 m^{2}+9 n^{2}+2 \operatorname{lm}-6 m n+3 n l\right)$

Question 5

- Factorise the following:
(i) $x^{3}+8 y^{3}+6 x y-1$
(ii) $l^{3}-8 m^{3}-27 n^{3}-18 \operatorname{lmn}$
Solution:
(i) $x^{3}+8 y^{3}+6 x y-1=x^{3}+(2 y)^{3}+(-1)^{3}-3(x)(2 y)(-1)$ $=(x+2 y-1)\left(x^{2}+4 y^{2}+1-2 x y+2 y+x\right)$
(ii) $l^{3}-8 m^{3}-27 n^{3}-18 l m n=l^{3}+(-2 m)^{3}+(-3 n)^{3}-3$ (l) $\{-2 m)(-3 n)$ $=(1-2 m-3 n)\left(1^{2}+(-2 m)^{2}+(-3 n)^{3}-1 \times-2 m-(-2 m \times-3 n)-(-3 n \times 1)\right)$ $=(1-2 m-3 n)\left(1^{2}+4 m^{2}+9 n^{2}+2 l m-6 m n+3 n l\right)$