Factorise the following, using the identity (a^2-b^2 )=(a-b)(a+b)… - Vidyadip

Question

Factorise the following, using the identity $\left(a^2-b^2\right)=(a-b)(a+b)$.
(i) $4 x^2-25 y^2$
(ii) $\frac{2 p^2}{25}-32 q^2$
(iii) $\frac{x^3 y}{9}-\frac{x y^3}{16}$
(iv) $16 x^4-81$

✓

Answer

(i) We have, $4 x^2-25 y^2=(2 x)^2-(5 y)^2$
$
=(2 x-5 y)(2 x+5 y)
$
(ii) We have,
$
\begin{aligned}
\frac{2 p^2}{25}-32 q^2 & =2\left(\frac{p^2}{25}-16 q^2\right)=2\left[\left(\frac{p}{5}\right)^2-(4 q)^2\right] \\
& =2\left(\frac{p}{5}-4 q\right)\left(\frac{p}{5}+4 q\right)
\end{aligned}
$
(iii) We have,
$
\begin{aligned}
\frac{x^3 y}{9}-\frac{x y^3}{16} & =x y\left(\frac{x^2}{9}-\frac{y^2}{16}\right)=x y\left[\left(\frac{x}{3}\right)^2-\left(\frac{y}{4}\right)^2\right] \\
& =x y\left(\frac{x}{3}-\frac{y}{4}\right)\left(\frac{x}{3}+\frac{y}{4}\right)
\end{aligned}
$
(iv) We have,
$
\begin{aligned}
16 x^4-81 & =\left(4 x^2\right)^2-(9)^2 \\
& =\left(4 x^2-9\right)\left(4 x^2+9\right) \\
& =\left[(2 x)^2-(3)^2\right]\left(4 x^2+9\right) \\
& =(2 x-3)(2 x+3)\left(4 x^2+9\right)
\end{aligned}
$

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