Exercise 3.2 - Chapter 3 Algebra 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

$\operatorname{Ex} 3.2$
Question 1.
Find the GCD of the given polynomials
(i) $x^{4}+3 x^{3}-x-3, x^{3}+x^{2}-5 x+3$
(ii) $x^{4}-1, x^{3}-11 x^{2}+x-11$
(iii) $3 x^{4}+6 x^{3}-12 x^{4}-24 x, 4 x^{4}+14 x^{3}+8 x^{2}-8 x$
(iv) $3 x^{3}+3 x^{2}+3 x+3,6 x^{3}+12 x^{2}+6 x+12$
Solution:
$x^{4}+3 x^{3}-x-3, x^{3}+x^{2}-5 x+3$
Let $f(x)=x^{4}+3 x^{3}-x-3$
$g(x)=x^{3}+x^{2}-5 x+3$

Note that 3 is not a divisor of $g(x)$. Now dividing $g(x)=x^{3}+x^{2}-5 x+3$ by the new remainder $x^{2}$ $+2 x-3$ (leaving the constant factor 3 ) we get

Here we get zero remainder
G.C.D of $\left(x^{4}+3 x^{3}-x-3\right),\left(x^{3}+x^{2}-5 x+3\right)$ is $\left(x^{2}+2 x-3\right)$
(ii) $x^{4}-1, x^{3}-11 x^{2}+x-11$

(iii) $3 x^{4}+6 x^{3}-12 x^{2}-24 x, 4 x^{4}+14 x^{3}+8 x^{2}-8 x$ $4 x^{4}+14 x^{3}+8 x^{2}-8 x=2\left(2 x^{4}+7 x^{3}+4 x^{2}-4 x\right)$
Let us divide
$\left(2 x^{4}+7 x^{3}+4 x^{2}+4 x\right)$ by $x^{4}+2 x^{3}-4 x^{2}-8 x$

$\therefore \mathrm{x}^{3}+4 \mathrm{x}^{2}+4 \mathrm{x}$ is the G.C.D of $3 \mathrm{x}^{4}+6 \mathrm{x}^{3}-12 \mathrm{x}^{2}-24 \mathrm{x}, 4 \mathrm{x}^{4}+14 \mathrm{x}^{3}+8 \mathrm{x}^{2}-8 \mathrm{x}$
$\therefore$ Ans $x\left(x^{2}+4 x+4\right)$
(iv) $f(x)=3 x^{3}+3 x^{2}+3 x+3=3\left(x^{3}+x^{2}+x+1\right)$
$\begin{aligned}
&g(x)=6 x^{3}+12 x^{2}+6 x+12 \\
&=6\left(x^{3}+2 x^{2}+x+2\right) \\
&=2 \times 3\left(x^{3}+2 x^{2}+x+2\right)
\end{aligned}$
$\begin{aligned}
&f(x) \Rightarrow x^{3}+x^{2}+x+1 \\
&\frac{g(x)}{6} \Rightarrow x^{3}+2 x^{2}+x+2
\end{aligned}$

Question 2.
Find the LCM of the given expressions,
(i) $4 x^{2} y, 8 x^{3} y^{2}$
(ii) $-9 \mathrm{a}^{3} \mathrm{~b}^{2}, 12 \mathrm{a}^{2} \mathrm{~b}^{2} \mathrm{c}$
(iii) $16 \mathrm{~m},-12 \mathrm{~m}^{2} \mathrm{n}^{2}, 8 \mathrm{n}^{2}$
(iv) $p^{2}-3 p+2, p^{2}-4$
(v) $2 x^{2}-5 x-3,4 x^{2}-36$
(vi) $\left(2 x^{2}-3 x y\right)^{2},(4 x-6 y)^{3}, 8 x^{3}-27 y^{3}$
Solution:
(i) $4 x^{2} y, 8 x^{3} y^{2}$
$4 x^{2} y=2 \times 2 x^{2} y$
$8 x^{3} y^{2}=\underline{2} \times \underline{2} \times \underline{2} x^{3} y^{2}$
L.C.M. $=2 \times 2 \times 2 x^{3} y^{2}$
$=8 x^{3} y^{2}$
(ii) $-9 a^{3} b^{2}=\underline{-3} \times \underline{3} a^{3} b^{2}$
$12 \mathrm{a}^{2} \mathrm{~b}^{2} \mathrm{c}=\underline{2} \times \underline{3} \times 2 \mathrm{a}^{2} \mathrm{~b}^{2} \mathrm{c}$
L.C.M. $=-3 \times 3 \times 2 \times 2 a^{3} b^{2} c$
$=-36 a^{3} b^{2} c$

$\begin{aligned}
&\text { (iii) } 16 \mathrm{~m},-12 \mathrm{~m}^{2} \mathrm{n}^{2}, 8 \mathrm{n}^{2} \\
&16 \mathrm{~m}=\underline{2} \times 2 \times 2 \times 2 \times \mathrm{m} \\
&-12 \mathrm{~m}^{2} \mathrm{n}^{2}=\underline{-2} \times \underline{2} \times 3 \times \mathrm{m}^{2} \mathrm{n}^{2} \\
&8 \mathrm{n}^{2}=2 \times 2 \times 2 \times \mathrm{n}^{2} \\
&\text { L.C.M. }=-2 \times 2 \times 2 \times 2 \times 3 \mathrm{~m}^{2} \mathrm{n}^{2} \\
&=-48 \mathrm{~m}^{2} \mathrm{n}^{2}
\end{aligned}$

(iv) $p^{2}-3 p+2, p^{2}-4$
$\begin{aligned}
p^{2}-3 p+2 &=(p-2)(p-1) \\
p^{2}-4 &=(p+2)(p-2) \\
\text { L.C.M. } &=(p-2)(p+2)(p-1)
\end{aligned}$

$\begin{aligned}
&\text { (v) } 2 x^{2}-5 x-3,4 x^{2}-36 \\
&2 x^{2}-5 x-3=(\underline{x-3})(2 x+1) \\
&4 x^{2}-36=4(x+3)(\underline{x-3}) \\
&\text { L.C.M. }=4(x+3)(x-3)(2 x+1) \\
&\text { (vi) }\left(2 x^{2}-3 x y\right)^{2}=(x(2 x-3 y))^{2} \\
&(4 x-6 y)^{3}=(2(2 x-3 y))^{3} \\
&8 x^{3}-27 y^{3}=(2 x)^{3}-(3 y)^{3} \\
&=(2 x-3 y)\left(4 x^{2}+6 x y+9 y^{2}\right) \\
&\text { L.C.M. }=2^{3} \times x^{2}(2 x-3 y)^{3}\left(4 x^{2}+6 x y+9 y^{2}\right)
\end{aligned}$