Question

factorise 4 x square + 9 y square + 16 z square + 12 x y - 24 Y Z - 16 X Z by using suitable identity

Answer 1

4x²+9y²+16z²+12xy-24yz-16xz
= (2x)²+(3y)²+(-4z)²+2[(2x)(3y)+(3y)(-4z)+(-4z)(2x)]
=(2x+3y-4z)²

Answer 2

Answer:

$4x^{2}+9y^{2}+16z^{2}+12xy-24yz-16xz\\=(2x+3y-4z)(2x+3y-4z)$

Step-by-step explanation:

$Given ,\: 4x^{2}+9y^{2}+16z^{2}+12xy-24yz-16xz$

/* we know the algebraic identity:

$\boxed {a^{2}+b^{2}+c^{2}+2xy+2yz+2zx=(a+b+c)^{2}}$

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

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

Therefore,

$4x^{2}+9y^{2}+16z^{2}+12xy-24yz-16xz\\=(2x+3y-4z)(2x+3y-4z)$

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