Question

Factorise: $\sqrt{ab}(x^2+1)+x(a+b)$

• A. $(\sqrt{b}x-\sqrt{a})(\sqrt{a}+\sqrt{b}x)$
• B. $(\sqrt{b}x+\sqrt{a})(\sqrt{a}x+\sqrt{b})$
• C. $(\sqrt{b}-\sqrt{a}x)(a-\sqrt{b}x)$
• D. $(\sqrt{b}+\sqrt{a}x)(\sqrt{a}+\sqrt{b}x)$

Answer 1

The correct option is B $(\sqrt{b}x+\sqrt{a})(\sqrt{a}x+\sqrt{b})$

We have to factorise:
$\sqrt{ab}(x^2+1)+x(a+b)$
$=\sqrt{ab}{x}^2+\sqrt{ab}+ax+bx$
$=\sqrt{ab}{x}^2++ax+bx+\sqrt{ab}$
Also,
$\sqrt{ab}\times \sqrt{ab}=ab$
$=1\times ab$
$=a\times b$
and
$a=\sqrt{a}\times \sqrt{a},b=\sqrt{b}\times \sqrt{b}$
Hence,
$=\sqrt{ab}{x}^2+ax+bx+\sqrt{ab}$
$=\sqrt{a}\times \sqrt{b}{x}^2+\sqrt{a}\times \sqrt{a}x+\sqrt{b}\times \sqrt{b}x+\sqrt{a}\times \sqrt{b}$
$=\sqrt{a}x(\sqrt{b}x+\sqrt{a})+\sqrt{b}(\sqrt{b}x+\sqrt{a})$
$=(\sqrt{b}x+\sqrt{a})(\sqrt{a}x+\sqrt{b})$
Hence, the correct answer is option (b).