Question

Resolve into factors: $9x^4+4y^4+11x^2{y}^2$

Answer 1

In the given expression $9x^4+4y^4+11x^2{y}^2$, add and subtract $12x^2{y}^2$ to make it a perfect square as shown below:

$9x^4+4y^4+11x^2{y}^2=(9x^4+4y^4+12x^2{y}^2)+11x^2{y}^2-12x^2{y}^2=\left[\right(3x^2{)}^2+(2y^2{)}^2+2(3\left)\right(2\left)x^2{y}^2\right]-x^2{y}^2$

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

(using the identity $(a+b{)}^2=a^2+b^2+2ab$

We also know the identity $a^2-b^2=(a+b)(a-b)$, therefore,

Using the above identity, the expression $(3x^2+2y^2{)}^2-x^2{y}^2$can be factorised as follows:

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

Hence, $9x^4+4y^4+11x^2{y}^2=(3x^2+2y^2+xy)(3x^2+2y^2-xy)$