Answer
$13-i\sqrt{2}$
Work Step by Step
Using $(a+b)(c+d)=ac+(ad+bc)+bd$ or the FOIL Method, the given expression, $
\left(3\sqrt{2}+i\right)\left(2\sqrt{2}-i\right),$ is
\begin{align*}
&
3\sqrt{2}\left(2\sqrt{2}\right)+\left(3\sqrt{2}\left(-i\right)+i\left(2\sqrt{2}\right)\right)+i\left(-i\right)
\\&=
6\left(\sqrt{2}\right)^2+\left(-3i\sqrt{2}+2i\sqrt{2}\right)-i^2
\\&=
6\left(2\right)-i\sqrt{2}-i^2
\\&=
12-i\sqrt{2}-i^2
.\end{align*}
Since $i^2=-1,$ the expression above is equivalent to
\begin{align*}
&
12-i\sqrt{2}-(-1)
\\&=
12-i\sqrt{2}+1
\\&=
(12+1)-i\sqrt{2}
\\&=
13-i\sqrt{2}
.\end{align*}
Hence, the expression $
\left(3\sqrt{2}+i\right)\left(2\sqrt{2}-i\right)
$ simplifies to $
13-i\sqrt{2}
$.
