Exercise 3.4 - Chapter 3 Algebra 8th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $3.4$
Question 1.
Factorise the following by taking out the common factor
(i) $18 x y-12 y z$
Answer:
$18 x y-12 y z=(2 \times 3 \times 3 \times y \times x)-(2 \times 2 \times 3 \times y \times z)$
Taking out the common factors 2,3 , y, we get $=2 \times 3 \times y(3 x-2 z)=6 y(3 x-2 z)$
(ii) $9 \mathrm{x}^{5} \mathrm{y}^{3}+6 \mathrm{x}^{3} \mathrm{y}^{2}-18 \mathrm{x}^{2} \mathrm{y}$
Answer:
$\begin{aligned}
&9 x^{5}+6 x^{3} y^{2}-18 x^{2} y=\left(3 \times 3 \times x^{2} \times x^{3} \times y \times y\right)+\left(2 \times 3 \times x^{2} \times x \times y \times y\right)-(2 \times 3 \times \\
&\left.3 \times x^{2} \times y\right)
\end{aligned}$
Taking out the common factors $3, x^{2}, y$, we get
$\begin{aligned}
&=3 \times x^{2} \times y\left(3 x^{3} y^{2}+2 x y-6\right) \\
&=3 x^{2} y\left(3 x^{3} y^{2}+2 x y-6\right)
\end{aligned}$
(iii) $x(b-2 c)+y(b-2 c)$
Answer:
Taking out the binomial factor $(b-2 c)$ from each term, we have $=(b-2 c)(x+y)$
(iv) $(a x+a y)+(b x+b y)$
Answer:
Taking at ' $a$ ' from the first term and ' $b$ ' from the second term we have $(a x+a y)+(b x+b y)=a(x+y)+b(x+y)$
Now taking out the binomial factor $(x+y)$ from each term $=(x+y)(a+b)$

(v) $2 x^{2}(4 x-1)-4 x+1$
Answer:
Taking out $-1$ from last two terms
$2 x^{2}(4 x-1)-4 x+1=2 x^{2}(4 x-1)-1(4 x-1)$
Taking out the binomial factor $4 x-1$, we get
$=(4 x-1)\left(2 x^{2}-1\right)$
(vi) $3 y(x-2)^{2}-2(2-x)$
Answer:
$3 y(x-2)^{2}-2(2-x)=3 y(x-2)(x-2)-2(-1)(x-2)$
$[\because$ Taking out $-1$ from $2-x]$
$=3 y(x-2)(x-2)+2(x-2)$
Taking out the binomial factor $x-2$ from each term, we get
$=(x-2)[3 y(x-2)+2]$
(vii) $6 x y-4 y^{2}+12 x y-2 y z x$
Answer:
$=6 x y+12 x y-4 y^{2}-2 y z x[\because$ Addition is commutative $]$
$=(6 \times \mathrm{x} \times \mathrm{y})+(2 \times 6 \times \mathrm{x} \times \mathrm{y})+(-1)(2)(2) \mathrm{y}+\mathrm{y})+((-1)(2)(\mathrm{y})(\mathrm{z})(\mathrm{x}))$
Taking out $6 x x x y$ from first two terms and $(-1) \times 2 \times y$ from last two terms we get
$=6 \times \mathrm{x} \times \mathrm{y}(1+2)+(-1)(2) \mathrm{y}[2 \mathrm{y}+\mathrm{zx}]$
$=6 \times y(3)-2 y(2 y+z x)$
$=(2 \times 3 \times 3 \times x \times y)-2 x y(2 y+z x)$
Taking out $2 y$ from two terms
$\begin{aligned}
&=2 y(9 x-(2 y+z x)) \\
&=2 y(9 x-2 y-x z)
\end{aligned}$

(viii) $a^{3}-3 a^{2}+a-3$
Answer:
$a^{2}-3 a^{2}+a-3=a^{2}(a-3)+1(a-3)$ [:Groupingthetermssuitably] $=(a-3)\left(a^{2}+1\right)$
(ix) $3 y^{3}-48 y$
Answer:
$3 y^{2}-48 y=3 \times y \times y^{2}-3 \times 16 \times y$
Taking out $3 \times y$
$=3 y\left(y^{2}-16\right)=3 y\left(y^{2}-4^{2}\right)$
Comparing $\mathrm{y}^{2}-4^{2}$ with $\mathrm{a}^{2}-\mathrm{b}^{2}$
$a=y, b=4$
$\begin{aligned}
&a^{2}-b^{2}=(a+b)(a-b) \\
&y^{2}-4^{2}=(y+4)(y-4) \\
&\therefore 3 y\left(y^{2}-16\right)=3 y(y+4)(y-4)
\end{aligned}$
(x) $a b^{2}-b c^{2}-a b+c^{2}$
Answer:
$a b^{2}-b c^{2}-a b+c^{2}$
Grouping suitably
$a b^{2}-b c^{2}-a b+c^{2}=b\left(a b-c^{2}\right)-1\left(a b-c^{2}\right)$
Taking out the binomial factor $a b-c^{2}=\left(a b-c^{2}\right)(b-1)$

Question $2 .$
Factorise the following expressions
(i) $x^{2}+14 x+49$
Answer:
$x^{2}+14 x+49=x^{2}+14 x+72$
Comparing with $\mathrm{a}^{2}+2 \mathrm{ab}+\mathrm{b}^{2}=(\mathrm{a}+\mathrm{b})^{2}$ we have $\mathrm{a}=\mathrm{x}$ and $\mathrm{b}=7$
$\Rightarrow x^{2}+2(x)(7)+7^{2}=(x+7)^{2}$
$\therefore x^{2}+14 x+49=(x+7)^{2}$
(ii) $\mathrm{y}^{2}-10 \mathrm{y}+25$
Answer:
$y^{2}-10 y+25=y^{2}-10 y+5^{2}$
Comparing with $\mathrm{a}^{2}-2 \mathrm{ab}+\mathrm{b}^{2}=(\mathrm{a}-\mathrm{b})^{2}$ we get $\mathrm{a}=\mathrm{y} ; \mathrm{b}=5$
$\begin{aligned}
&\Rightarrow y^{2}-2(y)(5)+5^{2}=(y-5)^{2} \\
&\therefore y^{2}-10 y+25=(y-5)^{2}
\end{aligned}$
(iii) $c^{2}-4 c-12$
Answer:
This is of the form $a x^{2}+b x+c$
Where a $=1, b=-4 \mathrm{c}=-12, \mathrm{x}=\mathrm{c}$
Now the product ac $=1 \times-12=-12$ and the sum $b=-4$

$\therefore$ The middle term $-4 \mathrm{c}$ can be written as $2 \mathrm{c}-6 \mathrm{c}$
$\begin{aligned}
&\therefore c^{2}-4 c-12=c^{2}+2 c-6 c-12 \\
&=c(c+2)-6(c+2)
\end{aligned}$

Taking out $(\mathrm{c}+2)$
$begin{aligned}
&\Rightarrow(c+2)(c-6) \\
&\therefore c^{2}-4 c-12=(c+2)(c-6)
\end{aligned}$
(iv) $\mathrm{m}^{2}+\mathrm{m}-72$
Answer:
$m^{2}+m-72$
This is of the form $a x+b x+c$ where $\mathrm{a}=1, \mathrm{~b}=1, \mathrm{c}=-72$

Product a $\times \mathrm{c}=1 \times-72=-72$
Sum b=1
The middle term $m$ can be written as $9 m-8 m$
$\begin{aligned}
&\mathrm{m}^{2}+\mathrm{m}-72=\mathrm{m}^{2}+9 \mathrm{~m}-8 \mathrm{~m}-72 \\
&=\mathrm{m}(\mathrm{m}+9)-8(\mathrm{~m}+9)
\end{aligned}$

Taking out $(\mathrm{m}+9)$
$=(\mathrm{m}+9)(\mathrm{m}-8)$
$\therefore \mathrm{m}^{2}+\mathrm{m}-72=(\mathrm{m}+9)(\mathrm{m}-8)$
(v) $4 x^{2}-8 x+3$
Answer:
$4 x^{2}-8 x+3$
This is of the form $a x^{2}+b x+c$ with $a=4 b=-8 c=3$
Product ac $=4 \times 3=12$
sum $b=-8$

The middle term can be written as $-8 x=-2 x-6 x$
$\begin{aligned}
&4 x^{2}-8 x+3=4 x^{2}-2 x-6 x+3 \\
&=2 x(2 x-1)-3(2 x-1) \\
&=(2 x-1)(2 x-3)
\end{aligned}$
$4 x^{2}-8 x+3=(2 x-1)(2 x-3)$

Question $3 .$
Factorise the following expressions using $(a+b)^{3}=a^{3}+3 a^{2} b+3 a b^{2}+b^{3}$ identity
(i) $64 \mathrm{x}^{3}+144 \mathrm{x}^{2}+108 \mathrm{x}+27$
(ii) $27 \mathrm{p}^{3}+54 \mathrm{p}^{3} \mathrm{q}+36 \mathrm{pq}^{2}+8 \mathrm{q}^{3}$
Answer:
(i) $64 x^{3}+144 x^{2}+108 x+27$
$\begin{aligned}
&=(4 x)^{3}+3(4 x)^{2}(3)+3(4)(3)^{2}+3^{3} \\
&=(4 x+3)^{3}
\end{aligned}$

$\begin{aligned}
&\text { (ii) } 27 p^{3}+54 p^{3} q+36 p q^{2}+8 q^{3} \\
&=(3 p)^{3}+3(3 p)^{2}(2 q)+3(3 p)(2 q)^{2}+(2 q)^{3} \\
&=(3 p+2 q)^{3}
\end{aligned}$

Question $4 .$
Factorise the following expressions using $(a-b)^{3}=a^{3}-3 a^{2} b+3 a b^{2}-b^{3}$ identity
(i) $y^{3}-18 y^{2}+108 y-216$
(ii) $8 m^{3}-60 m^{2} n+150 m n^{2}-125 n^{3}$
Answer:
$\begin{aligned}
&\text { (i) } y^{3}-18 y^{2}+108 y-216 \\
&=y^{3}-3 y^{2}(6)+3(6)^{2} y-6^{3} \\
&=(y-6)^{3}
\end{aligned}$
(ii) $8 m^{3}-60 m^{2} n+150 m n^{2}-125 n^{3}$
$=(2 m)^{3}-3(2 m)^{2}(5)+3(2 m)(5 n)^{2}-(5 n)^{3}$
$=(2 m-5 n)^{3}$