Question

Factorise :
$\text{(i)}{x}^2+8x+16$
$\text{(ii)}4a^2–4a+1$

Answer 1

We have,
$\text{(i)}{x}^2+8x+16=x^2+2\times x\times 4+4^2$
$=(x+4{)}^2\text{[Using}:a^2+2ab+b^2=(a+b{)}^2]$
$=(x+4)(x+4)$
$\text{(ii)}4a^2–4a+1=(2a{)}^2–2\times 2a\times 1+(1{)}^2$
$=(2a–1{)}^2\text{[Using}:a^2–2ab+b^2=(a–b{)}^2]$
$=(2a–1)(2a–1)$