Question

Factorise: 2x² + y² + 2z²-2√2xy +2√2yz - 4x2.​

Answer 1

Given :- Factorise: 2x² + y² + 2z²-2√2xy +2√2yz - 4xz .

Solution :-

→ 2x² + y² + 2z²-2√2xy +2√2yz - 4xz

as we can see that, negative sign is with (2√2xy) and 4xz) . so, x must be negative. (common in both)

then,

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

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

using a² + b² + c² + 2ab + 2bc + 2ca = (a + b + c)²

→ (-√2x + y + √2z)²

→ (y + √2z - √2x)(y + √2z - √2x)(y + √2z - √2x) (Ans.)

Learn more :-

Let a, b and c be non-zero real numbers satisfying (a³)/(b³ + c³) + (b³)/(c³ + a³) + (c³)/(a³ + b³)

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if a²+ab+b²=25

b²+bc+c²=49

c²+ca+a²=64

Then, find the value of

(a+b+c)² - 100 = __

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