Answer
The standard form of the expression $3\sqrt{-7}\left( 2\sqrt{-8} \right)$ is $-12\sqrt{14}$.
Work Step by Step
Consider the expression,$3\sqrt{-7}\left( 2\sqrt{-8} \right)$
Express the square roots of negative numbers in terms of $i$.
$\begin{align}
& 3\sqrt{-7}\left( 2\sqrt{-8} \right)=3i\sqrt{7}\left( 2i\sqrt{8} \right) \\
& =6{{i}^{2}}\sqrt{7}\left( \sqrt{8} \right)
\end{align}$
Replace the value ${{i}^{2}}=-1$ and make the factors of the radicals.
$\begin{align}
& 3\sqrt{-7}\left( 2\sqrt{-8} \right)=6\left( -1 \right)\sqrt{7}\left( \sqrt{4\cdot 2} \right) \\
& =-6\sqrt{7}\left( 2\sqrt{2} \right) \\
& =-12\sqrt{7}\cdot \sqrt{2}
\end{align}$
Use the property $\sqrt{a}\cdot \sqrt{b}=\sqrt{ab}$.
$\begin{align}
& 3\sqrt{-7}\left( 2\sqrt{-8} \right)=-12\sqrt{7\cdot 2} \\
& =-12\sqrt{14}
\end{align}$
Therefore, the standard form of the expression $3\sqrt{-7}\left( 2\sqrt{-8} \right)$ is $-12\sqrt{14}$.
