Answer
$\dfrac{f(x)}{g(x)}=
2x^2-10x+30-\dfrac{89}{x+3}$;
$x=-3$ is not in the domain
Work Step by Step
With the given functions, $f(x)=
2x^3-4x^2+1
,$ and $g(x)=
x+3
,$ using the long division below, then
\begin{array}{l}\require{cancel}
\dfrac{f(x)}{g(x)}
=\dfrac{
2x^3-4x^2+1
}{
x+3
}
\\\\
\dfrac{f(x)}{g(x)}=
2x^2-10x+30-\dfrac{89}{x+3}.\end{array}
If $x=
-3
,$ then the denominator, $
g(x)=x+3
,$ becomes zero. Since denominators cannot be zero, then $
x=-3
$ is not in the domain of $\dfrac{f(x)}{g(x)}.$
