Question

Factorise:
$x^2+2y^2+8z^2+2\sqrt{2}xy-8yz-4\sqrt{2}xz$

Answer 1

We have,
$x^2+2y^2+8z^2+2\sqrt{2}xy-8yz-4\sqrt{2}xz$
$=(x{)}^2+(\sqrt{2}y{)}^2+(-2\sqrt{2}z{)}^2+2(x\left)\right(\sqrt{2}y)+2(\sqrt{2}y\left)\right(-\sqrt{2}z)+2(x\left)\right(-2\sqrt{2}z)$
$[\because a^2+b^2+c^2+2ab+2bc+2ca=(a+b+c{)}^2]$
$=(x+\sqrt{2}y-2\sqrt{2}z{)}^2$
$=(x+\sqrt{2}y-2\sqrt{2}z)(x+\sqrt{2}y-2\sqrt{2}z)$