How do you multiply  (a^2sqrt8) / (sqrt16a^6)?

Mar 31, 2016

$\frac{\sqrt{2}}{2 {a}^{4}}$

Explanation:

$1$. Start by simplifying all square roots. For $\sqrt{8}$, use perfect squares to simplify.

$\frac{{a}^{2} \sqrt{8}}{\sqrt{16} {a}^{6}}$

$= \frac{{a}^{2} \sqrt{4 \times 2}}{4 {a}^{6}}$

$= \frac{{a}^{2} \cdot 2 \sqrt{2}}{4 {a}^{6}}$

$= \frac{2 {a}^{2} \sqrt{2}}{4 {a}^{6}}$

$2$. Factor out $2$ from the numerator and denominator.

$= \frac{2 \left({a}^{2} \sqrt{2}\right)}{2 \left(2 {a}^{6}\right)}$

$= \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \left({a}^{2} \sqrt{2}\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \left(2 {a}^{6}\right)}$

$= \frac{{a}^{2} \sqrt{2}}{2 {a}^{6}}$

$3$. Use the exponent quotient law, ${\textcolor{p u r p \le}{b}}^{\textcolor{red}{m}} \div {\textcolor{p u r p \le}{b}}^{\textcolor{b l u e}{n}} = {\textcolor{p u r p \le}{b}}^{\textcolor{red}{m} - \textcolor{b l u e}{n}}$, to simplify $\frac{{a}^{2}}{{a}^{6}}$. Since the power in the denominator has a larger exponent, calculate $\frac{1}{a} ^ \left(6 - 2\right)$, instead of $\frac{{a}^{2 - 6}}{1}$, which would save you a step.

$= \frac{\sqrt{2}}{2 {a}^{6 - 2}}$

$= \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \frac{\sqrt{2}}{2 {a}^{4}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$