Question

Factorise: $3x^4+12y^4$

Answer 1

In the given equation $3x^4+12y^4$ or $3(x^4+4y^4)$, add and subtract $4x^2{y}^2$ to make it a perfect square as shown below:

$4x^4+81y^4=3\left[\right(x^4+4y^4+4x^2{y}^2)-4x^2{y}^2]=3\left\{\right[(x^2{)}^2+(2y^2{)}^2+2\left(x^2\right)\left(2y^2\right)]-4x^2{y}^2\}=3\left[\right(x^2+2y^2{)}^2-4x^2{y}^2]$

(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 equation $(x^2+2y^2{)}^2-4x^2{y}^2$can be factorised as follows:

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

Hence, $3x^4+12y^4=3\left[\right(x^2+2y^2+2xy\left)\right(x^2+2y^2-2xy\left)\right]$