## Algebra 2 (1st Edition)

$\displaystyle \frac{8}{\sqrt{3}}=\frac{8\sqrt{3}}{3}$
$\displaystyle \frac{8}{\sqrt{3}}\qquad$ ...rationalize the denominator by multyplying both the numerator and the denominator with $\sqrt{3}$. $=\displaystyle \frac{8\cdot\sqrt{3}}{\sqrt{3}\cdot\sqrt{3}}\qquad$ ...apply the Product Property $\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}$ in the denominator. $=\displaystyle \frac{8\sqrt{3}}{\sqrt{3\cdot 3}}\qquad$ ...simplify $=\displaystyle \frac{8\sqrt{3}}{\sqrt{9}}\qquad$ ...evaluate the denominator ($\sqrt{9}=3$) $=\displaystyle \frac{8\sqrt{3}}{3}$