Question

Factorise $4{(x-y)}^2-9{(x-y)}^4$.

Answer 1

Step I - 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:

$4{(x-y)}^2-9{(x-y)}^4$ = ${\left[2\right(x-y\left)\right]}^2-[3{(x-y)}^2{]}^2$

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

Step II - Here, $(x-y)$ is the common factor from each term. So we get,

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

= ${(x-y)}^2(2-3x+3y)(2+3x-3y)$

Therefore factors of $4{(x-y)}^2-9{(x-y)}^4$ =${(x-y)}^2(2-3x+3y)(2+3x-3y)$.