## Algebra 2 (1st Edition)

$\sqrt{108}\qquad$ ...rewrite 18 as a product of two factors so that one factor is a perfect square. ($108=36\cdot 3$) $=\sqrt{36\cdot 3}\qquad$ ...use the Product Property $\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}$ $=\sqrt{36}\cdot\sqrt{3}\qquad$ ...evaluate $\sqrt{36}$ ($\sqrt{36}=6$) $=6\sqrt{3}$