# How do you simplify sqrt(2x)(sqrt(8x)-sqrt32)?

Oct 4, 2017

See a solution process below:

#### Explanation:

First, rewrite we can expand the expression by multiplying each term in parenthesis by the term outside the parenthesis:

$\sqrt{\textcolor{red}{2 x}} \left(\sqrt{\textcolor{b l u e}{8 x}} - \sqrt{\textcolor{b l u e}{32}}\right) \implies$

$\left(\sqrt{\textcolor{red}{2 x}} \cdot \sqrt{\textcolor{b l u e}{8 x}}\right) - \left(\sqrt{\textcolor{red}{2 x}} \cdot \sqrt{\textcolor{b l u e}{32}}\right)$

We can next use this rule for radicals to execute the two multiplication operations:

$\sqrt{\textcolor{red}{a}} \cdot \sqrt{\textcolor{b l u e}{b}} = \sqrt{\textcolor{red}{a} \cdot \textcolor{b l u e}{b}}$

$\left(\sqrt{\textcolor{red}{2 x}} \cdot \sqrt{\textcolor{b l u e}{8 x}}\right) - \left(\sqrt{\textcolor{red}{2 x}} \cdot \sqrt{\textcolor{b l u e}{32}}\right) \implies$

$\sqrt{\textcolor{red}{2 x} \cdot \textcolor{b l u e}{8 x}} - \sqrt{\textcolor{red}{2 x} \cdot \textcolor{b l u e}{32}} \implies$

$\sqrt{16 {x}^{2}} - \sqrt{64 x} \implies$

$4 x - \sqrt{\textcolor{red}{64} \cdot \textcolor{b l u e}{x}} \implies$

$4 x - \left(\sqrt{\textcolor{red}{64}} \cdot \sqrt{\textcolor{b l u e}{x}}\right) \implies$

$4 x - 8 \sqrt{x}$