Question

Factorise $x^2{y}^2+x{y}^3-x^3{y}^4-x^2{y}^5$.

Answer 1

Step I - Here $x{y}^2$ is the common factors from each term. So we get,

$x^2{y}^2+x{y}^3-x^3{y}^4-x^2{y}^5$ = $x{y}^2(x+y-x^2{y}^2-x{y}^3)$

Step II - Now rearranging the terms in the brackets.

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

Step III - Taking the common factors from first two and last two terms in the brackets. We get,

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

Here $(1-x{y}^2)$ is the common factor from each term in the bracket.

So, $x{y}^2[x(1-x{y}^2)+y(1-x{y}^2\left)\right]$ = $x{y}^2(1-x{y}^2)(x+y)$

Hence, factors of $x^2{y}^2+x{y}^3-x^3{y}^4-x^2{y}^5$ = $x{y}^2(1-x{y}^2)(x+y)$ .