Question

Factorise
$2{x}^{2} + 2 \sqrt{6} xy + 3y {}^{2}$
By Factorization of the perfect square polynomials method.

Most satisfying answer will mark as brainliest. ​

Answer 1

Step-by-step explanation:

2x^2+2√6xy+3y^2

=(√2x) ^2+2(√2)(√3y) +(√3y) ^2

= (√2x + √3y) ^2

=(√2x+√3y) (√2x+√3y)

Answer 2

Answer:

$= > 2x ^{2} + 2 \sqrt{6} xy + 3y ^{2} = 0$

$= > ( \sqrt{2x} ) ^{2} + 2 \times \sqrt{2x} \times \sqrt{3y} + \sqrt{3y} ^{2} = 0$

(a+b)²=a²+2ab+b²

$= >({ \sqrt{2x} + \sqrt{3y} }) ^{2} = 0$

$= > ({\sqrt{2x} + \sqrt{3y} })( \sqrt{2x} + \sqrt{3y} ) = 0$