Answer
The simplified form of $\left( 2-i\sqrt{3} \right)\left( 6+i\sqrt{3} \right)$ is $15-4i\sqrt{3}$.
Work Step by Step
Consider the expression.
$\left( 2-i\sqrt{3} \right)\left( 6+i\sqrt{3} \right)$
Apply the FOIL method.
$\begin{align}
& \left( 2-i\sqrt{3} \right)\left( 6+i\sqrt{3} \right)=2\cdot 6+2\cdot i\sqrt{3}-i\sqrt{3}\cdot 6-i\sqrt{3}\cdot i\sqrt{3} \\
& =12+-i4\sqrt{3}-{{i}^{2}}{{\left( \sqrt{3} \right)}^{2}}
\end{align}$
Apply the radical rule, and substitute ${{i}^{2}}=-1$ .
$\begin{align}
& \left( 2-i\sqrt{3} \right)\left( 6+i\sqrt{3} \right)=12-i4\sqrt{3}-\left( -1 \right)3 \\
& =12-i4\sqrt{3}+3 \\
& =15-4i\sqrt{3}
\end{align}$
