## Precalculus (6th Edition) Blitzer

$5$; $6\sqrt3$
Simplifying the radical gives $\sqrt{3} + \sqrt{25\cdot 3} \\=\sqrt{3} + \sqrt{5^2 \cdot 3} \\=\sqrt{3} + \sqrt{5^2} \cdot \sqrt{3} \\=\sqrt{3} + \underline{5}\sqrt{3}.$ Use the rule $ac+bc=(a+b)c$ to obtain $=(1+5)\sqrt{3} \\=\underline{6\sqrt{3}}.$ Thus, the missing expressions are: $5$ and $6\sqrt{3}.$