Question

Factorise: $x^2-\frac{9}{16}$

Answer 1

Given that: $x^2-\frac{9}{16}$

$=x^2-\frac{3^2}{4^2}$

$=x^2-{\left(\frac{3}{4}\right)}^2$

$=(x+\frac{3}{4})(x-\frac{3}{4})$, since $a^2-b^2=(a+b)(a-b)$

The factors of $x^2-\frac{9}{16}$ are $(x+\frac{3}{4})(x-\frac{3}{4})$.