Question

Factorise $a^2-1\frac{9}{16}{b}^2$.

Answer 1

As we know the algebraic identity,

$a^2-b^2=(a+b)(a-b)$

So, it can be factorized by using this identity, we can write:

$a^2-1\frac{9}{16}{b}^2$ $=$${\left(a\right)}^2-{\left(\frac{5}{4}b\right)}^2$

$=$$\left(a+\frac{5}{4}b\right)\left(a-\frac{5}{4}b\right)$

Therefore factors of $a^2-1\frac{9}{16}{b}^2$ $=$$\left(a+\frac{5}{4}b\right)\left(a-\frac{5}{4}b\right)$.