Question

Factorise: 

(i) 4x ² + 9y ² + 16z ² + 12xy – 24yz – 16xz 

(ii) 2x ² + y ² + 8z ² – 2√2 xy + 4√2 yz – 8xz

Answer 1

(i) It is known that, 

(x + y + z)² = x2 + y² + z² + 2xy + 2yz + 2zx

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

= (2x + 3y - 4z)² = (2x + 3y - 4z)(2x + 3y - 4z)

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

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

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