# How do you simplify  (3 sqrt(x^3)) times (4 + 2 sqrt(xy))?

Jan 8, 2018

See a solution process below:

#### Explanation:

First, multiply each term within the right parenthesis by the term on the left:

$\left(\textcolor{red}{3 \sqrt{{x}^{3}}}\right) \times \left(4 + 2 \sqrt{x y}\right) \implies$

$\left(\textcolor{red}{3 \sqrt{{x}^{3}}} \times 4\right) + \left(\textcolor{red}{3 \sqrt{{x}^{3}}} \times 2 \sqrt{x y}\right) \implies$

$12 \sqrt{{x}^{3}} + 6 \sqrt{{x}^{3}} \sqrt{x y}$

Next, we can use this rule to combine the radicals in the term on the right:

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

$12 \sqrt{{x}^{3}} + 6 \sqrt{\textcolor{red}{{x}^{3}}} \sqrt{\textcolor{b l u e}{x y}} \implies$

$12 \sqrt{{x}^{3}} + 6 \sqrt{\textcolor{red}{{x}^{3}} \cdot \textcolor{b l u e}{x y}} \implies$

$12 \sqrt{{x}^{3}} + 6 \sqrt{{x}^{4} y}$

Then, we can use the opposite of the above rule to reduce the radicals:

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

$12 \sqrt{{x}^{3}} + 6 \sqrt{{x}^{4} y} \implies$

$12 \sqrt{\textcolor{red}{{x}^{2}} \cdot \textcolor{b l u e}{x}} + 6 \sqrt{\textcolor{red}{{x}^{4}} \cdot \textcolor{b l u e}{y}} \implies$

$12 \sqrt{\textcolor{red}{{x}^{2}}} \sqrt{\textcolor{b l u e}{x}} + 6 \sqrt{\textcolor{red}{{x}^{4}}} \sqrt{\textcolor{b l u e}{y}} \implies$

$12 x \sqrt{x} + 6 {x}^{2} \sqrt{y}$