# How do you multiply 4\sqrt { 15a } \cdot 4\sqrt { 3a }?

Mar 11, 2018

See a solution process below:

#### Explanation:

First, rewrite the expression as:

$4 \cdot 4 \cdot \sqrt{15 a} \sqrt{3 a} \implies$

$16 \sqrt{15 a} \sqrt{3 a}$

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

$16 \sqrt{\textcolor{red}{15 a}} \cdot \sqrt{\textcolor{b l u e}{3 a}} \implies$

$16 \sqrt{\textcolor{red}{15 a} \cdot \textcolor{b l u e}{3 a}} \implies$

$16 \sqrt{45 {a}^{2}}$

Then, rewrite the term in the radical as:

$16 \sqrt{9 {a}^{2} \cdot 5}$

Now, use this rule of radicals to complete the simplification:

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

$16 \sqrt{\textcolor{red}{9 {a}^{2}} \cdot \textcolor{b l u e}{5}} \implies$

$16 \sqrt{\textcolor{red}{9 {a}^{2}}} \cdot \sqrt{\textcolor{b l u e}{5}} \implies$

$16 \cdot 3 a \cdot \sqrt{\textcolor{b l u e}{5}} \implies$

$48 a \sqrt{5}$