Question

Resolve into factors:
$4x^4+25y^4+10x^2{y}^2$

Answer 1

We know the identity $a^4+b^4+a^2{b}^2=(a^2+b^2+ab)(a^2+b^2-ab)$

Using the above identity, the equation $4x^4+25y^4+10x^2{y}^2$ can be factorised as follows:

$4x^4+25y^4+10x^2{y}^2=(\sqrt{2}x{)}^4+(\sqrt{5}y{)}^4+\left(\sqrt{2}x{)}^2\right(\sqrt{5}y{)}^2$

$=\left[\right(\sqrt{2}x{)}^2+(\sqrt{5}y{)}^2+(\sqrt{2}x\left)\right(\sqrt{5}y\left)\right]\left[\right(\sqrt{2}x{)}^2+(\sqrt{5}y{)}^2+(\sqrt{2}x\left)\right(\sqrt{5}y\left)\right]=(2x^2+5y^2+\sqrt{10}xy)$ $(2x^2+5y^2-\sqrt{10}xy)$

Hence, $4x^4+25y^4+10x^2{y}^2=(2x^2+5y^2+\sqrt{10}xy)(2x^2+5y^2-\sqrt{10}xy)$