Answer
$b^2+6b+3$
Work Step by Step
Given \begin{equation}
\left(b^4+6 b^3-18 b-9\right) \div\left(b^2-3\right).
\end{equation} Use long division. $$
\begin{array}{r}
b^2+6b+3\phantom{)} \\
b^2-3{\overline{\smash{\big)}\,b^4+6 b^3+0b^2-18 b-9\phantom{)}}}\\
\underline{-~\phantom{(}(b^4-3b^2)\phantom{-b)}}\\
0+6b^3+3b^2-18b\phantom{)}\\
\underline{-~\phantom{()}(6b^3-18b)}\\
0+3b^2-9\phantom{)}\\
\underline{-~\phantom{()}(3b^2-9)}\\
0\phantom{)}
\end{array}
$$ The solution is \begin{equation}
\frac{b^4+6 b^3-18 b-9}{b^2-3}=b^2+6b+3.
\end{equation}
