Question

Question 26
Factorise the following:
(i) $4x^2+20x+25$

(ii) $9y^2–66yz+121z^2$

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

Answer 1

$\left(i\right)4x^2+20x+25=(2x{)}^2+2\times 5\times 2x+(5{)}^2$
$=(2x+5{)}^2$ [Using the identity, $a^2+2ab+b^2=(a+b{)}^2$]

$\left(ii\right)9y^2–66yz+121z^2=(3y{)}^2–2\times 3y\times 11z+(11z{)}^2$
$=(3y–11z{)}^2$ [Using the identity, $a^2–2ab+b^2=(a–b{)}^2$]

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

[Using the identity,
$a^2–b^2=(a–b)(a+b)]$

$=(2x-x+\frac{1}{3}+\frac{1}{2})(2x+x+\frac{1}{3}-\frac{1}{2})$
$=(x+\frac{2+3}{6})(3x+\frac{2-3}{6})$
$=(x+\frac{5}{6})(3x-\frac{1}{6})$