Factorise the following


Factorise the following

(i) 9x2 +4y2 + 16z2 +12xy-16yz -24xz

(ii) 25x2 + 16y2 + 4z2 – 40xy + 16yz – 20xz

(iii) 16x2 + 4y2 + 9z2 – 16xy – 12yz + 24xz


(i) $9 x^{2}+4 y^{2}+16 z^{2}+12 x y-16 y z-24 x z$

$=(3 x)^{2}+(2 y)^{2}+(-4 z)^{2}+2(3 x)(2 y)+2(2 y)(-4 z)+2(-4 z)(3 x)$

[using identity, $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a$ ]

$=(3 x+2 y-4 z)^{2}=(3 x+2 y-4 z)(3 x+2 y-4 z)$

(ii) $25 x^{2}+16 y^{2}+4 z^{2}-40 x y+16 y z-20 x z$

$=(-5 x)^{2}+(4 y)^{2}+(2 z)^{2}+2(-5 x)(4 y)+2(4 y)(2 z)+2(2 z)(-5 x)$

[using identity, $\left.(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a\right]$

$=(-5 x+4 y+2 z)^{2}=(-5 x+4 y+2 z)(-5 x+4 y+2 z)$

(iii) $16 x^{2}+4 y^{2}+9 z^{2}-16 x y-12 y z+24 x z$

$=(4 x)^{2}+(-2 y)^{2}+(3 z)^{2}+2(4 x)(-2 y)+2(-2 y)(3 z)+2(4 x)(3 z)$

[using identity, $\left.(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a\right]$

$=(4 x-2 y+3 z)^{2}=(4 x-2 y+3 z)(4 x-2 y+3 z)$

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now