Answer
$3x^2+2x+4$
Work Step by Step
Given \begin{equation}
\left(3 x^4+2 x^3+22 x^2+12 x+24\right) \div\left(x^2+6\right)
\end{equation} Use long division.
$$
\begin{array}{r}
3x^2+2x+4\phantom{)} \\
x^2+6{\overline{\smash{\big)}\,3 x^4+2 x^3+22 x^2+12 x+24\phantom{)}}}\\
\underline{-~\phantom{(}(3x^4+18x^2)\phantom{-b)}}\\
0+2x^3+4x^2+12x\phantom{)}\\
\underline{-~\phantom{()}(2x^3+12x)}\\
0+4x^2+24\phantom{)}\\
\underline{-~\phantom{()}(4x^2+24)}\\
0\phantom{)}
\end{array}
$$ The solution is
\begin{equation}
\frac{3 x^4+2 x^3+22 x^2+12 x+24}{x^2+6}=3x^2+2x+4
\end{equation}
