Answer
$5\sqrt{3}-4\sqrt{2}$
Work Step by Step
Factor each radicand so that one factor is a perfect square:
$\sqrt{25\cdot3}-4\sqrt{9\cdot2}+2\sqrt{16\cdot2}$
Recall the property (pg. 367):
$\sqrt[n]{a}\cdot\sqrt[n]{b}=\sqrt[n]{ab}$ (if $\sqrt[n]{a}$ and $\sqrt[n]{b}$ are real numbers)
Applying this property, we get:
$\sqrt{25\cdot3}-4\sqrt{9\cdot2}+2\sqrt{16\cdot2}$
$=\sqrt{25}\cdot \sqrt{3}-4\sqrt{9}\cdot \sqrt{2}+2\sqrt{16}\cdot \sqrt{2}$
Recall that $5^2=25$, $3^2=9$ and $4^2=16$.
Thus, the expression above simplifies to:
$5\sqrt{3}-4\cdot 3\sqrt{2}+2\cdot4\sqrt{2}$
$=5\sqrt{3}-12\sqrt{2}+8\sqrt{2}$
$=5\sqrt{3}+(8-12)\sqrt{2}$
$=5\sqrt{3}-4\sqrt{2}$