Factorise the following

Question:

Factorise the following

(i) $4 x^{2}+20 x+25$

(ii) $9 y^{2}-66 y z+121 z^{2}$

(iii) $\left(2 x+\frac{1}{3}\right)^{2}-\left(x-\frac{1}{2}\right)^{2}$

Solution:

(i) $4 x^{2}+20 x+25=(2 x)^{2}+2 \times 2 x \times 5+(5)^{2}$

$=(2 x+5)^{2}$  [using identity, $a^{2}+2 a b+b^{2}=(a+b)^{2}$ ]

(ii) $9 y^{2}-66 y z+121 z^{2}=(3 y)^{2}-2 \times 3 y \times 11 z+(11 z)^{2}$

$=(3 y-11 z)^{2} \quad$ [using identity, $a^{2}-2 a b+b^{2}=(a-b)^{2}$ ]

(iii) $\left(2 x+\frac{1}{3}\right)^{2}-\left(x-\frac{1}{2}\right)^{2}=\left[\left(2 x+\frac{1}{3}\right)-\left(x-\frac{1}{2}\right)\right]\left[\left(2 x+\frac{1}{3}\right)+\left(x-\frac{1}{2}\right)\right]$

[using identity, $a^{2}-b^{2}=(a-b)(a+b)$ ]

$=\left(2 x-x+\frac{1}{3}+\frac{1}{2}\right)\left(2 x+x+\frac{1}{3}-\frac{1}{2}\right)$

$=\left(x+\frac{2+3}{6}\right)\left(3 x+\frac{2-3}{6}\right)$

$=\left(x+\frac{5}{6}\right)\left(3 x-\frac{1}{6}\right)$

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