Answer
$4b^2-3b+5$
Work Step by Step
Given \begin{equation}
\left(4 b^4-3 b^3-23 b^2+21 b-35\right) \div\left(b^2-7\right)
\end{equation} Use long division.
$$
\begin{array}{r}
4b^2-3b+5\phantom{)} \\
b^2-7{\overline{\smash{\big)}\,4 b^4-3 b^3-23 b^2+21 b-35\phantom{)}}}\\
\underline{-~\phantom{(}(4b^4-28b^2)\phantom{-b)}}\\
0-3 b^3+5b^2+21b\phantom{)}\\
\underline{-~\phantom{()}(-3b^3+21b)}\\
0+5b^2-35\phantom{)}\\
\underline{-~\phantom{()}(5b^2-35)}\\
0\phantom{)}
\end{array}
$$ The solution is
\begin{equation}
\frac{4b^4-3 b^3-23b^2+21b-35}{b^2-7}=4b^2-3b+5
\end{equation}
