Factorise: <br/> <br/>(i) $4 x^{2}+9 y^{2}+16 z^{2}+12 x y-24 y z-16 x z$<br/> <br/>(ii) $2 x^{2}+y^{2}+8 z^{2}-2 \sqrt{2} x y+4 \sqrt{2} y z-8 x z$
Solution:
It is known that,
$(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 z x$
(i) $4 x^{2}+9 y^{2}+16 z^{2}+12 x y-24 y z-16 x z$
$=(2 x)^{2}+(3 y)^{2}+(-4 z)^{2}+2(2 x)(3 y)+2(3 y)(-4 z)+2(2 x)(-4 z)$
$=(2 x+3 y-4 z)^{2}$
$=(2 x+3 y-4 z)(2 x+3 y-4 z)$
(ii) $2 x^{2}+y^{2}+8 z^{2}-2 \sqrt{2} x y+4 \sqrt{2} y z-8 x z$
$=(-\sqrt{2} x)^{2}+(y)^{2}+(2 \sqrt{2} z)^{2}+2(-\sqrt{2} x)(y)+2(y)(2 \sqrt{2} z)+2(-\sqrt{2} x)(2 \sqrt{2} z)$
$=(-\sqrt{2} x+y+2 \sqrt{2} z)^{2}$
$=(-\sqrt{2} x+y+2 \sqrt{2} z)(-\sqrt{2} x+y+2 \sqrt{2} z)$
It is known that,
$(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 z x$
(i) $4 x^{2}+9 y^{2}+16 z^{2}+12 x y-24 y z-16 x z$
$=(2 x)^{2}+(3 y)^{2}+(-4 z)^{2}+2(2 x)(3 y)+2(3 y)(-4 z)+2(2 x)(-4 z)$
$=(2 x+3 y-4 z)^{2}$
$=(2 x+3 y-4 z)(2 x+3 y-4 z)$
(ii) $2 x^{2}+y^{2}+8 z^{2}-2 \sqrt{2} x y+4 \sqrt{2} y z-8 x z$
$=(-\sqrt{2} x)^{2}+(y)^{2}+(2 \sqrt{2} z)^{2}+2(-\sqrt{2} x)(y)+2(y)(2 \sqrt{2} z)+2(-\sqrt{2} x)(2 \sqrt{2} z)$
$=(-\sqrt{2} x+y+2 \sqrt{2} z)^{2}$
$=(-\sqrt{2} x+y+2 \sqrt{2} z)(-\sqrt{2} x+y+2 \sqrt{2} z)$