Question

factorise 2x^2+y^2+8z^2-2√2 xy + 4√2 yz-8xz
please tell fast please ​

Answer 1

Answer:

(-√2x + y + 2√2z)(- √2x + y + 2√2z)

Step-by-step explanation:

Given

2x^2 + y^2 + 8z^2 – 2√2xy + 4√2yz – 8xz

Using identity,

(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx

We can say that,

x^2 + y^2 + z^2 + 2xy + 2yz + 2zx = (x + y + z)^2

2x^2 + y^2 + 8z^2 – 2√2xy + 4√2yz – 8xz

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

= (-√2x + y + 2√2z)^2

= (-√2x + y + 2√2z)(- √2x + y + 2√2z)

Answer 2

ANSWER :

Given 2x2 + y2 + 8z2 – 2√2xy + 4√2yz – 8xz

Using identity,

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

We can say that,

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

2x2 + y2 + 8z2 – 2√2xy + 4√2yz – 8xz

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

= (-√2x + y + 2√2z)2

= (-√2x + y + 2√2z)(- √2x + y + 2√2z)