Question

factorise (i) 4x² + 9y² + 16z² + 12xy - 24yz - 16xz , (ii) 2x² + y² + 8z² - 2√2 xy + 4√2 yz - 8 xz

Answer 1

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Answer 2

$\mathbb{ ANSWER }$ :

(i) 4x² + 9y² + 16z² + 12xy - 24yz - 16xz

[ $\therefore$ (x+y+z) (x² + y² + x² + 2xy + 2yz + 2xz ]

= (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 )

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(ii) 2x² + y² + 8z² - 2√2 xy + 4√2 yz - 8 xz

[ $\therefore$ x+y+z = x² + y² + z² + 2xy + 2yz + 2xz ]

= (-√2x)² + (y)² + (2√2z)² + 2(-√2x*y) + 2(y*2√2) + 2(2√2z*-√2x)

= ( -√2x + y + 2√2z )²

= ( -√2x + y + 2√2z ) ( -√2x + y + 2√2z )