Question

Factorise:

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

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

Answer 1

(i) We have,

9x² + 4y² + 16z² – 12xy – 16yz + 24xz

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

[∵ a² + b² + c² + 2ab + 2bc + 2ca = (a + b + c)²]

= (3x – 2y + 4z)²

= (3x – 2y + 4z)(3x – 2y + 4z)

(ii) We have,

x² + 2y² + 8z² + 2√2xy – 8yz – 4√2xz 

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

[∵ a² + b² + c² + 2ab + 2bc + 2ca = (a + b + c)²]

= (x + √2y – 2√2z)²

= (x + √2y – 2√2z)(x + √2y – 2√2z)