Exercise 3.6 - Chapter 3 Algebra 9th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

$\operatorname{Ex} 3.6$
Question $1 .$
Factorise the following:
(i) $x^{2}+10 x+24$
(ii) $z^{2}+4 z-12$
(iii) $p^{2}-6 p-16$
(iv) $\mathrm{t}^{2}+72-17 \mathrm{t}$
(v) $y^{2}-16 y-80$
(vi) $a^{2}+10 a-600$
Solution:
(i) $x^{2}+10 x+24$
$x^{2}+10 x+24=x^{2}+6 x+4 x+24$

= x(x + 6) + 4 (x + 6) = (x + 6) (x + 4)

(ii) $z^{2}+4 z-12$
$z^{2}+4 z-12=z^{2}+6 z-2 z-12$
$\begin{aligned}
&=z(z+6)-2(z+6) \\
&=(z+6)(z-2)
\end{aligned}$

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

$\begin{aligned}
&=\mathrm{p}(\mathrm{p}-8)+2(\mathrm{p}-8) \\
&=(\mathrm{p}-8)(\mathrm{p}+2) \\
&\text { (iv) } \mathrm{t}^{2}+72-17 \mathrm{t} \\
&\mathrm{t}^{2}+72-17 \mathrm{t}=\mathrm{t}^{2}-17 \mathrm{t}+72 \\
&=\mathrm{t}^{2}-9 \mathrm{t}-8 \mathrm{t}+72 \\
&72 \\
&=\mathrm{t}(\mathrm{t}-9)-8(\mathrm{t}-9) \\
&=(\mathrm{t}-9)(\mathrm{t}-8) \\
&\text { (v) } \mathrm{y}^{2}-16 \mathrm{y}-80 \\
&\mathrm{y}^{2}-16 \mathrm{y}-80=\mathrm{y}^{2}-20 \mathrm{y}+4 \mathrm{y}-80 \\
&80 \\
&\quad-16 \\
&=\mathrm{y}(\mathrm{y}-20)+4(\mathrm{y}-20) \\
&=(\mathrm{y}-20)(\mathrm{y}+4) \\
&\text { (vi) } \mathrm{a}^{2}+10 \mathrm{a}-600 \\
&\mathrm{a}^{2}+10 \mathrm{a}-600=\mathrm{a}^{2}+30 \mathrm{a}-20 \mathrm{a}-600
\end{aligned}$

$=a(a+30)-20(a+30)$
$=(a+30)(a-20)$

Question 2.
Factorise the following
(i) $2 a^{2}+9 a+10$
(ii) $5 \mathrm{x}^{2}-29 \mathrm{xy}-42 \mathrm{y}^{2}$
(iii) $9-18 x+18 x^{2}$
(iv) $6 x^{2}+16 x y+8 y^{2}$
(v) $12 x^{2}+36 x^{2} y+27 y^{2} x^{2}$
(vi) $(a+b)^{2}+9(a+b)+18$
Solution:
(i) $2 a^{2}+9 a+10$
$2 \mathrm{a}^{2}+9 \mathrm{a}+10=2 \mathrm{a}^{2}+4 \mathrm{a}+5 \mathrm{a}+10$

$\begin{aligned}
&=2 a(a+2)+5(a+2) \\
&=(a+2)(2 a+5)
\end{aligned}$
(ii) $5 x^{2}-29 x y-42 y^{2}$ $5 x^{2}-35 x y+6 x y-42 y^{2}$

$\begin{aligned}
&=5 x(x-7)+6 y(x-7) \\
&=(x-7)(5 x+6 y) \\
&\text { (iii) } 9-18 x+8 x^{2} \\
&=8 x^{2}-18 x+9 \\
&=8 x^{2}-6 x-12 x+9
\end{aligned}$

$\begin{aligned}
&=2 x(4 x-3)-3(4 x-3) \\
&=(4 x-3)(2 x-3)
\end{aligned}$
(iv) $6 x^{2}+16 x y+8 y^{2}$
$=2\left(3 x^{2}+8 x y+4 y^{2}\right)$
$=2\left(3 x^{2}+8 x y+4 y^{2}\right)$
$\begin{aligned}
&=2(3 x(x+2 y)+2 y(x+2 y)) \\
&=2(x+2 y)(3 x+2 y)
\end{aligned}$
(v) $12 x^{2}+36 x^{2} y+27 y^{2} x^{2}$
$=27 y^{2} x^{2}+36 x^{2} y+12 x^{2}=3 x^{2}\left(9 y^{2}+12 y+4\right)$

$\begin{aligned}
&=3 x^{2}\left(9 y^{2}+6 y+6 y+4\right)=3 x^{2}(3 y(3 y+2)+2(3 y+2)) \\
&=3 x^{2}(3 y+2)(3 y+2)=3 x^{2}(3 y+2)(3 y+2) \\
&(v i)(a+b)^{2}+9(a+6)+18 \\
&=(a+b)^{2}+6(a+b)+3(a+b)+18 \\
\end{aligned}$

= ((a + 6) + 6) ((a + b) + 3) = (a + b + 6) (a + b + 3)

Question 3.

Factorise the following:
(i) $(p-q)^{2}-6(p-q)-16$
(ii) $\mathrm{m}^{2}+2 \mathrm{mn}-24 \mathrm{n}^{2}$
(iii) $\sqrt{5} a^{2}+2 a-3 \sqrt{5}$
(iv) $\mathrm{a}^{4}-3 \mathrm{a}^{2}+2$
(v) $8 m^{3}-2 m^{2} n-15 m^{2}$
(vi) $\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{2}{x y}$
Solution:
(i) $(p-q)^{2}-6(p-q)-16$
$=(p-q)^{2}-8(p-q)+2(p-q)-16$

$\begin{aligned}
&=(p-q)((p-q)-8)+2((p-q)-8) \\
&=(p-q-8)(p-q+2) \\
&\text { (ii) } m^{2}+2 m n-24 n^{2} \\
&=m^{2}+6 m n-4 m n-24 n^{2}
\end{aligned}$

$\begin{aligned}
&=m(m+6 n)-4 n(m+6 n) \\
&=(m+6 n)(m-4 n)
\end{aligned}$

(iii) $\sqrt{5} a^{2}+2 a-3 \sqrt{5}$
$\begin{aligned}
&=\sqrt{5} a^{2}+2 a-3 \sqrt{5} \\
&=\sqrt{5} a^{2}+5 a-3 a-3 \sqrt{5} \\
&=\sqrt{5} a(a+\sqrt{5})-3(a+\sqrt{5}) \\
&=(a+\sqrt{5})(\sqrt{5} a-3)
\end{aligned}$

(iv) $a^{4}-3 a^{2}+2$
$\begin{aligned}
&=a^{4}-2 a^{2}-1 a^{2}+2=a^{2}\left(a^{2}-2\right)-1\left(a^{2}-2\right) \\
&=\left(a^{2}-2\right)\left(a^{2}-1\right)=\left(a^{2}-2\right)(a+1)(a-1)
\end{aligned}$
(v) $8 m^{3}-2 m^{2} n-15 m n^{2}$
$\begin{aligned}
&=m\left(8 m^{2}-2 m n-15 n^{2}\right) \\
&=m\left(8 m^{2}-12 m n+10 m n-15 n^{2}\right) \\
&=m(4 m(2 m-3 n)+5 n(2 m-3 n) \\
&=m(4 m+5 n)(2 m-3 n)
\end{aligned}$

(vi) $\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{2}{x y}=\left(\frac{1}{x}\right)^{2}+2\left(\frac{1}{x}\right)\left(\frac{1}{y}\right)+\left(\frac{1}{y}\right)^{2}=\left(\frac{1}{x}+\frac{1}{y}\right)^{2}$