Answer
$3x\sqrt{2x}$
Work Step by Step
RECALL:
(i) For any non-negative real numbers $a$ and $b$,
$\sqrt{a} \cdot \sqrt{b} =\sqrt{ab}.$
(ii) For any non-negative real number $a$, $\sqrt{a^2}=a$.
Use rule (i) above to obtain:
$\sqrt{6x(3x^2)}
\\=\sqrt{18x^{2+1}}
\\=\sqrt{18x^3}.$
Factor the radicand so that one of the factors is a perfect square to obtain
$\sqrt{9x^2(2x)}
\\=\sqrt{(3x)^2(2x)}.$
Use rule (ii) above where $a=3x$ to obtain
$\\3x\sqrt{2x}.$
