## Intermediate Algebra for College Students (7th Edition)

$-10i\sqrt{3}$
Factor the radicand so that one factor $-1$ to obtain: $=-\sqrt{300(-1)} \\=-\sqrt{100(3)(-1)} \\=-\sqrt{10^2(3)(-1)}$ RECALL: (1) $\sqrt{abc} = \sqrt{a} \cdot \sqrt{b} \cdot \sqrt{c}$ (2) $\sqrt{-1} = i$ Use rule (1) above to obtain: $=-\sqrt{10^2} \cdot \sqrt{3} \cdot \sqrt{-1} \\=-10 \cdot \sqrt{3} \cdot \sqrt{-1} \\=-10\sqrt{3} \cdot \sqrt{-1}$ Use rule (2) above to obtain: $=-10\sqrt{3} \cdot i \\=-10i\sqrt{3}$