Answer
$4a^2-3a+7$
Work Step by Step
Given \begin{equation}
\left(4a^4-11a^3+29a^2-26a+28\right) \div\left(a^2-2a+4\right)
\end{equation} Use long division. $$
\begin{array}{r}
4a^2-3a+7\phantom{)} \\
a^2-2a+4{\overline{\smash{\big)}\,4a^4-11a^3+29a^2-26a+28\phantom{)}}}\\
\underline{-~\phantom{(}(4a^4-8a^3+16a^2)\phantom{-b)}}\\
0-3a^3+13a^2-26a\phantom{)}\\
\underline{-~\phantom{()}(-3a^3+6a^2-12a)}\\
0+7a^2-14a+28\phantom{)}\\
\underline{-~\phantom{()}(7a^2-14a+28)}\\
0\phantom{)}
\end{array}
$$ The solution is
\begin{equation}
\frac{4a^4-11a^3+29a^2-26a+28}{a^2-2a+4}=4a^2-3a+7
\end{equation}
