Answer
$x^2-3x+4$
Work Step by Step
Given \begin{equation}
\left(x^4-3 x^3+8 x^2-12 x+16\right) \div\left(x^2+4\right).
\end{equation}
Use long division. $$
\begin{array}{r}
x^2-3x+4\phantom{)} \\
x^2+4{\overline{\smash{\big)}\,x^4-3 x^3+8 x^2-12 x+16\phantom{)}}}\\
\underline{-~\phantom{(}(x^4+4x^2)\phantom{-b)}}\\
0-3x^3+4x^2-12x\phantom{)}\\
\underline{-~\phantom{()}(-3x^3-12x)}\\
0+4x^2+16\phantom{)}\\
\underline{-~\phantom{()}(4x^2+16)}\\
0\phantom{)}
\end{array}
$$ The solution is \begin{equation}
\frac{x^4-3 x^3+8 x^2-12 x+16}{x^2+4}= x^2-3x+4.
\end{equation}
