Exercise 3.9 - Chapter 3 Algebra 9th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex 3.9
Question 1.
Find the GCD for the following:
(i) $\mathrm{p}^{5}, \mathrm{p}^{11}, \mathrm{p}^{9}$
(ii) $4 \mathrm{x}^{3}, \mathrm{y}^{3}, \mathrm{z}^{3}$
(iii) $9 \mathrm{a}^{2} \mathrm{~b}^{2} \mathrm{c}^{3}, 15 \mathrm{a}^{3} \mathrm{~b}^{2} \mathrm{c}^{4}$
(iv) $64 x^{8}, 240 x^{6}$
(v) $a b^{2} c^{3}, a^{2} b^{3} c, a^{3} b c^{2}$
(vi) $35 \mathrm{x}^{5} \mathrm{y}^{3} \mathrm{z}^{4}, 49 \mathrm{x}^{2} \mathrm{yz}^{3}, 14 \mathrm{xy}^{2} \mathrm{z}^{2}$
(vii) $25 \mathrm{ab}^{3} \mathrm{c}, 100 \mathrm{a}^{2} \mathrm{bc}, 125 \mathrm{ab}$
(viii) 3 abc, $5 x y z, 7 p q r$
Solution:
(i) $\mathrm{p}^{5}, \mathrm{p}^{11}, \mathrm{p}^{9}$
$\begin{aligned}
&\mathrm{p}^{5}=\mathrm{p}^{5} \\
&\mathrm{p}^{9}=\mathrm{p}^{5} \times \mathrm{p}^{4} \\
&\mathrm{p}^{11}=\mathrm{p}^{5} \times \mathrm{p}^{6}
\end{aligned}$
G.C.D. of $\mathrm{p}^{5}, \mathrm{p}^{11}, \mathrm{p}^{9}=\mathrm{p}^{5}$
(ii) $4 \mathrm{x}^{3}, \mathrm{y}^{3}, \mathrm{z}^{3}$
$\begin{aligned}
&\text { G.C.D. of } 4 \mathrm{x}^{3}=1 \times 4 \mathrm{x}^{3} \\
&\mathrm{y}^{3}=\underline{1} \times \mathrm{y}^{3} \\
&\mathrm{z}^{3}=\underline{1} \times \mathrm{z}^{3} \\
&\therefore \text { G.C.D. }=1
\end{aligned}$

$\begin{aligned}
&\text { (iii) } 9 \mathrm{a}^{2} \mathrm{~b}^{2} \mathrm{c}^{3}, 15 \mathrm{a}^{3} \mathrm{~b}^{2} \mathrm{c}^{4} \\
&9 \mathrm{a}^{2} \mathrm{~b}^{2} \mathrm{c}^{3}=\underline{3} \times 3 \mathrm{a}^{2} \mathrm{~b}^{2} \mathrm{c}^{3} \\
&15 \mathrm{a}^{3} \mathrm{~b}^{2} \mathrm{c}^{4}=\underline{3} \times 5 \mathrm{a}^{3} \mathrm{~b}^{2} \mathrm{c}^{4} \\
&\text { G.C.D. }=9 \mathrm{a}^{2} \mathrm{~b}^{2} \mathrm{c}^{3}, 15 \mathrm{a}^{3} \mathrm{~b}^{2} \mathrm{c}^{4}=3 \times \mathrm{a}^{2} \mathrm{~b}^{2} \mathrm{c}^{3}=3 \mathrm{a}^{2} \mathrm{~b}^{2} \mathrm{c}^{3}
\end{aligned}$

(iv) $64 x^{8}, 240 x^{6}$
$$
\begin{aligned}
&64 \mathrm{x}^{8}=\underline{2} \times \underline{2} \times \underline{2} \times \underline{2} \times \underline{2} \times \underline{2} \times \mathrm{x}^{6} \times \mathrm{x}^{8} \\
&240 \mathrm{x}^{6}=\underline{2} \times \underline{2} \times \underline{2} \times \underline{2} \times 3 \times 5 \mathrm{x}^{6} \\
&\text { G.C.D. of } 64 \mathrm{x}^{8}, 240 \mathrm{x}^{6}=2 \times 2 \times 2 \times 2 \mathrm{x}^{6}=16 \mathrm{x}^{6}
\end{aligned}
$$
(v) $a b^{2} c^{3}, a^{2} b^{3} c, a^{3} b c^{2}$
$\mathrm{ab}^{2} \mathrm{c}^{3}=\underline{\mathrm{a}} \times \underline{\mathrm{b}} \times \mathrm{b} \times \underline{\mathrm{c}} \times \mathrm{c} \times \mathrm{c}$
$\mathrm{a}^{2} \mathrm{~b}^{3} \mathrm{c}=\underline{a} \times \mathrm{a} \times \mathrm{b} \times \underline{\mathrm{b}} \times \mathrm{b} \times \underline{\mathrm{c}}$
$\mathrm{a}^{3} \mathrm{bc}^{2}=\underline{a} \times \underline{a} \times a \times \underline{b} \times \underline{c} \times c$
G.C.D. of $\mathrm{ab}^{2} \mathrm{c}^{3}, \mathrm{a}^{2} \mathrm{~b}^{3} \mathrm{c}, \mathrm{a}^{3} \mathrm{bc}^{2}=\mathrm{a} \times \mathrm{b} \times \mathrm{c}=\mathrm{abc}$
(vi) $35 x^{5} y^{3} z^{4}, 49 x^{2} y^{3}, 14 x y^{2} z^{2}$
$\begin{aligned}
&35 \mathrm{x}^{5} \mathrm{y}^{3} \mathrm{z}^{4}=5 \times 7 \mathrm{x}^{5} \mathrm{y}^{3} \mathrm{z}^{4} \\
&49 \mathrm{x}^{2} \mathrm{yz}^{3}=7 \times 7 \mathrm{x}^{2} \mathrm{y} \mathrm{z}^{3} \\
&14 \mathrm{xy}^{2} \mathrm{z}^{2}=2 \times 7 \mathrm{x} \mathrm{y}^{2} \mathrm{z}^{2} \\
&\text { G.C.D. is }=7 \mathrm{xy} \mathrm{z}^{2}
\end{aligned}$
$\begin{aligned}
&\text { (vii) } 25 \mathrm{ab}^{3} \mathrm{c}, 100 \mathrm{a}^{2} \mathrm{bc}, 125 \mathrm{ab} \\
&25 \mathrm{ab}^{3} \mathrm{c}=\underline{5} \times \underline{5} \mathrm{ab}^{3} \mathrm{c} \\
&100 \mathrm{a}^{2} \mathrm{bc}=2 \times \underline{5} \times 2 \times \underline{5} \mathrm{a}^{2} \mathrm{bc} \\
&125 \mathrm{ab}=\underline{5} \times \underline{5} \times 5 \mathrm{ab} \\
&\text { G.C.D is }=5 \times 5 \mathrm{ab}=25 \mathrm{ab}
\end{aligned}$

(viii) 3 abc, $5 x y z, 7 p q r$
$3 \mathrm{abc}=\underline{1} \times \underline{3} \mathrm{abc}$
$5 \mathrm{xyz}=1 \times 5 \times \mathrm{yz}$
$7 \mathrm{pqr}=\underline{1} \times 7 \mathrm{pqr}$
G.C.D is 1

Question 2.

Find the GCD of the following
(i) $(2 x+5),(5 x+2)$
(ii) $a^{m+1}, a^{m+2}, a^{m+3}$
(iii) $2 \mathrm{a}^{2}+\mathrm{a}, 4 \mathrm{a}^{2}-1$
(iv) $3 \mathrm{a}^{2}, 5 \mathrm{~b}^{3}, 7 \mathrm{c}^{4}$
(v) $x^{4}-1, x^{2}-1$
(vi) $a^{3}-9 a x^{2},(a-3 x)^{2}$
Solution:

$\begin{aligned}
&\text { (i) }(2 x+5),(5 x+2)\\
&(2 \mathrm{x}+5)=\underline{1} \times(2 \mathrm{x}+5)\\
&(5 x+2)=\underline{1} \times(5 x+2)\\
&\therefore \text { G.C.D }=1\\
&\text { (ii) } \mathrm{a}^{\mathrm{m}+1}, \mathrm{a}^{\mathrm{m}+2}, \mathrm{a}^{\mathrm{m}+3}\\
&\mathrm{a}^{\mathrm{m}+1}=\underline{\mathrm{a}}^{\mathrm{m}} \times \underline{\mathrm{a}}^{l}\\
&a^{m+2}=\underline{a}^{m} \times \underline{a}^{1} \times a^{l}\\
&a^{m+3}=\underline{a}^{m} \times \underline{a}^{1} \times a^{1} \times a^{1}\\
&\text { G.C.G }=a^{m} \times a^{1}=a^{m+1}\\
&\text { (iii) } 2 \mathrm{a}^{2}+\mathrm{a}, 4 \mathrm{a}^{2}-1\\
&2 \mathrm{a}^{2}+\mathrm{a}=\mathrm{a}(2 \mathrm{a}+1)\\
&4 a^{2}-1=(2 a)^{2}-1^{2}=(2 a+1)(2 a-1)\\
&\text { G.C.D }=(2 a+1)\\
&\text { (iv) } 3 \mathrm{a}^{2}, 5 \mathrm{~b}^{3}, 7 \mathrm{c}^{4}\\
&3 \mathrm{a}^{2}=\underline{1} \times 3 \mathrm{a} \times \mathrm{a}\\
&5 \mathrm{~b}^{3}=\underline{1} \times 5 \mathrm{~b} \times \mathrm{b} \times \mathrm{b}\\
&7 \mathrm{c}^{4}=\underline{1} \times 7 \times \mathrm{c} \times \mathrm{c} \times \mathrm{c} \times \mathrm{c}\\
&\therefore \text { G.C.D }=1\\
&\text { (v) } x^{4}-1, x^{2}-1\\
&x^{4}-1=\left(x^{2}\right)^{2}-1^{2}=\left(x^{2}+1\right)\left(x^{2}-1\right)\\
&=\left(\mathrm{x}^{2}+1\right)\left(\mathrm{x}^{2}-1^{2}\right)=\left(\mathrm{x}^{2}+1\right)(\underline{\mathrm{x}+1})(\underline{\mathrm{x}-1})\\
&x^{2}-1=x^{2}-1^{2}=(x+1)(x-1)\\
&\text { G.C.D }=(x+1)(x-1)=x^{2}-1
\end{aligned}$

$\begin{aligned}
&\text { (vi) } a^{3}-9 a x^{2},(a-3 x)^{2} \\
&a^{3}-9 a x^{2}=a\left(a^{2}-(3 x)^{2}\right)=a(a+3 x)(\underline{a-3 x}) \\
&(a-3 x)^{2}=(\underline{a-3 x})(a-3 x) \\
&\text { G.C.D }=(a-3 x)
\end{aligned}$