Exercise 3.3 - Chapter 3 Algebra 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Question $1 .$
Find the LCM and GCD for the following and verify that $f(x) \times g(x)=L C M \times G C D$.
(i) $21 x^{2} y, 35 x y^{2}$
(ii) $\left(x^{3}-1\right)(x+1), x^{3}+1$
(ii) $\left(x^{3}-1\right)(x+1),\left(x^{3}-1\right)$
(iii) $\left(x^{2} y+x y^{2}\right),\left(x^{2}+x y\right)$
Solution:
(i) $f(x)=21 x^{2} y=3 \times 7 x^{2} y$
$g(x)=35 x y^{2}=7 \times 5 x y^{2}$
G.C.D. $=7 x y$
L.C.M. $=7 \times 3 \times 5 \times x^{2} y^{2}=105 x^{2} \times y^{2}$
L.C.M $\times$ G.C.D $=f(x) \times g(x)$
$105 x^{2} y^{2} \times 7 x y=21 x^{2} y \times 35 x y^{2}$
$735 x^{3} y^{3}=735 x^{3} y^{3}$
Hence verified.
(ii) $\left(x^{3}-1\right)(x+1)=(x-1)\left(x^{2}+x+1\right)(x+1)$
$x^{3}+1=(x+1)\left(x^{2}-x+1\right)$
G.C.D $=(x+1)$
L.C.M $=(x-1)(x+1)\left(x^{2}+x+1\right)\left(x^{2}-x+1\right)$
$\therefore \mathrm{L}, \mathrm{C}, \mathrm{M}, \times \mathrm{G}, \mathrm{C}, \mathrm{D}=\mathrm{f}(\mathrm{x}) \times \mathrm{g}(\mathrm{x})$
$(x-1)(x+1)\left(x^{2}+x+1\right)\left(x^{2}-x+1\right)=(x-1)$
$\left(x^{2}+x+1\right) \times(x+1)\left(x^{2}-x+1\right)$
$\left(x^{3}-1\right)(x+1)\left(x^{3}+1\right)=\left(x^{3}-1\right)(x+1)\left(x^{3}+1\right)$
$\therefore$ Hence verified.

(iii) $f(x)=x^{2} y+x y^{2}=x y(x+y)$
$g(x)=x^{2}+x y=x(x+y)$
L.C.M. $=x y(x+y)$
G.C.D. $=x(x+y)$
To verify:
L.C.M. $\times$ G.C.D. $=x y(x+y) \times(x+y)$
$=x^{2} y(x+y)^{2} \ldots \ldots \ldots .(1)$
$f(x) \times g(x)=\left(x^{2} y+x y^{2}\right)\left(x^{2}+x y\right)$
$=x^{2} y(x+y)^{2} \ldots \ldots \ldots \ldots . .(2)$
$\therefore$ L.C.M. $\times$ G.C.D $=f(x) \times g(x)$.
Hence verified.

Question $2 .$
Find the LCM of each pair of the following polynomials
(i) $\mathrm{a}^{2}+4 \mathrm{a}-12, \mathrm{a}^{2}-5 \mathrm{a}+6$ whose GCD is $\mathrm{a}-2$
(ii) $x^{4}-27 a^{3} x,(x-3 a)^{2}$ whose GCD is $(x-3 a)$
Solution:
(i) $f(x)=a^{2}+4 a-12=(a+6)(a-2)$
$g(x)=a^{2}-5 a+6=(a-3)(a-2)$ G.C.D. $=(a-2)$ L.C.M. $=(a-2)(a-3)(a+6)$
L.C.M. $=\frac{f(x) \times g(x)}{\text { G.C.D }}$
$=\frac{(a+b)(a-2) \times(a-3)(a-2)}{(a-2)}$
$=(a-2)(a-3)(a+6)$
$\mathrm{L}, \mathrm{C}, \mathrm{M}_{+}=(a-3)\left(a^{2}+4 a-12\right)$
(ii) $f(x)=x^{4}-27 a^{3} x=x\left(x^{3}-(3 a)^{3}\right)$
$g(x)=(x-3 a)^{2}$
G.C.D - $(x-3 a)$
L.C.M. $\times$ G.C.D $=f(x) \times g(x)$
L. C.M $=\frac{x\left(x^{3}-(3 a)^{3}\right) \times(x-3 a)^{2}}{(x-3 a)}$
L.C.M $=x\left(x^{3}-(3 a)^{3}\right) \cdot(x-3 a)$
$=x(x-3 a)^{2}\left(x^{2}+3 a x+9 a^{2}\right)$
 

Question $3 .$
Find the GCD of each pair of the following polynomials
(i) $12\left(x^{4}-x^{3}\right), 8\left(x^{4}-3 x^{3}+2 x^{2}\right)$ whose LCM is $24 x^{3}(x-1)(x-2)$
(ii) $\left(x^{3}+y^{3}\right),\left(x^{4}+x^{2} y^{2}+y^{4}\right)$ whose LCM is $\left(x^{3}+y^{3}\right)\left(x^{2}+x y+y^{2}\right)$
Solution:
(i) $f(x)=12\left(x^{4}-x^{3}\right)$
$g(x)=8\left(x^{4}-3 x^{3}+2 x^{2}\right)$
L.C.M $=24 \mathrm{x}^{3}(\mathrm{x}-1)(\mathrm{x}-2)$
G.C.D. $=\frac{f(x) \times g(x)}{\text { L.C.M. }}$
$=\frac{12\left(x^{4}-x^{3}\right) \times 8\left(x^{4}-3 x^{3}+2 x^{2}\right)}{24 x^{3}(x-1)(x-2)}$
$=\frac{4 x^{3}(x-1) x^{2}\left(x^{2}-3 x+2\right)}{x^{3}(x-1)(x-2)}$

$\begin{aligned}
&=\frac{4 x^{2}(x-2)(x-1)}{(x-2)} \\
&=4 x^{2}(x-1)
\end{aligned}$
(ii) $\left(x^{3}+y^{3}\right),\left(x^{4}+x^{2} y^{2}+y^{4}\right)$
$\begin{aligned}
&\text { L.C.M. }=\left(x^{3}+y^{3}\right)\left(x^{2}+x y+y^{2}\right) \\
&\text { G.C.D }=\frac{f(x)(g(x))}{\text { L.C.M. }} \\
&=\frac{\left(x^{3}+y^{3}\right)\left(x^{4}+x^{2} y^{2}+y^{4}\right)}{\left(x^{3}+y^{5}\right)\left(x^{2}+x y+y^{2}\right)} \\
&=\frac{\left(x^{2}-x y+y^{2}\right)\left(x^{2}+x y+y^{2}\right)}{\left(x^{2}+y y+y^{2}\right)}=\left(x^{2}-x y+y^{2}\right)
\end{aligned}$

Question $4 .$
Given the LCM and GCD of the two polynomials $p(x)$ and $q(x)$ find the unknown polynomial in the following table

Solution: