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

Question

Factorise the following, using the identity $\left(a^2-2 a b+b^2\right)=(a-b)^2$.
(i) $y^2-14 y+49$
(ii) $\frac{x^2}{4}-2 x+4$
(iii) $a^2 y^3-2 a b y^2+b^2 y$
(iv) $9 y^2-4 x y+\frac{4 x^2}{9}$

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Answer

(i) $\begin{array}{l} y^2-14 y+49=y^2-2 \times 7 \times y+7^2 \\ =(y-7)^2=(y-7)(y-7) \\ \text { (ii) } \begin{aligned} \frac{x^2}{4}-2 x+4 & =\left(\frac{x}{2}\right)^2-2 \times \frac{x}{2} \times 2+2^2 \\ & =\left(\frac{x}{2}-2\right)^2=\left(\frac{x}{2}-2\right)\left(\frac{x}{2}-2\right)\end{aligned}\end{array}$
(iii)
$
\begin{aligned}
a^2 y^3-2 a b y^2+b^2 y & =y\left(a^2 y^2-2 a b y+b^2\right) \\
& =y\left[(a y)^2-2 \times(a y) \times b+b^2\right] \\
& =y(a y-b)^2=y(a y-b)(a y-b)
\end{aligned}
$
(iv)
$
\begin{aligned}
9 y^2-4 x y+\frac{4 x^2}{9} & =(3 y)^2-2 \times 3 y \times \frac{2 x}{3}+\left(\frac{2 x}{3}\right)^2 \\
& =\left(3 y-\frac{2 x}{3}\right)^2 \\
& =\left(3 y-\frac{2 x}{3}\right)\left(3 y-\frac{2 x}{3}\right)
\end{aligned}
$

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