Question

$\sqrt{6}{x}^{2}y+(2x+\sqrt{6})y+3xy$ is equal to

• A. $y(\sqrt{3}x-\sqrt{2})(\sqrt{3}x+2)$
• B. $y(\sqrt{2}x+\sqrt{3})(\sqrt{3}x+\sqrt{2})$
• C. $y(\sqrt{6}x+1)(1+\sqrt{3})$
• D. $y(\sqrt{3}{x}^2+2)(2+\sqrt{3})$

Answer 1

The correct option is B $y(\sqrt{2}x+\sqrt{3})(\sqrt{3}x+\sqrt{2})$

$\sqrt{6}{x}^{2}y+(2x+\sqrt{6})y+3xy$ can be written as:
Taking $y$ common
$=y[\sqrt{6}{x}^2+2x+\sqrt{6}+3x]$
$=y[\sqrt{6}{x}^2+2x+3x+\sqrt{6}]$
$=y[\sqrt{2}x(\sqrt{3}x+\sqrt{2})+\sqrt{3}(\sqrt{3}x+\sqrt{2})]$
$\{\sqrt{6}=\sqrt{2}\times \sqrt{3},2=\sqrt{2}\times \sqrt{2},3=\sqrt{3}\times \sqrt{3}\}$
$=y(\sqrt{3}x+\sqrt{2})(\sqrt{2}x+\sqrt{3})$
So, the correct answer is option (b).