Factorise: 2x^2+y^2+8z^2-2√2xy+4√2yz -8xz

Question

Factorise: \( 2 x^{2}+y^{2}+8 z^{2}-2 \sqrt{2} x y+4 \sqrt{2} y z-8 x z \)

Answer 1

\(2 x^{2}+y^{2}+8 z^{2}-2 \sqrt{2} x y+4 \sqrt{2} y z-8 x z \)

\(=(\sqrt{2})^{2} x^{2}+y^{2}+(2 \sqrt{2})^{2} z^{2}-2 \sqrt{2} x y+4 \sqrt{2} y z-8 x z\)

\(=(\sqrt{2} x)^{2}+y^{2}+(2 \sqrt{2} z)^{2}-2 \sqrt{2} x y+4 \sqrt{2} y z-8 x z \)

\(=(-\sqrt{2} x)^{2}+y^{2}+(2 \sqrt{2} z)^{2}-2 \sqrt{2} x y+4 \sqrt{2} y z-8 x z \)

\( =(-\sqrt{2} x)^{2}+y^{2}+(2 \sqrt{2} z)^{2}+2(-\sqrt{2} x)(y)+2(y)(2 \sqrt{2} z)+2(-\sqrt{2} x)(2 \sqrt{2} z)\)

Using (a+b+c)² = a² + b² + c² + 2ab + 2bc + 2ac 

Putting alpha = \(- \sqrt2 x,\,b=y,\,c=2\sqrt2x\)

\(= (-\sqrt 2 x+y +2 \sqrt 2z)^2\)

\(= (-\sqrt {2}x +y +2 \sqrt 2 z)(- \sqrt 2x +y +2 \sqrt 2 z)\)