Answer
$\frac{t^3+3 t^2-12 t+16}{t+3}=t^2-12+\frac{52}{t+3}$
Work Step by Step
Given \begin{equation}
\begin{aligned}
\left(t^3+3 t^2-12 t+16\right) \div(t+3).
\end{aligned}
\end{equation} Use synthetic division.
\begin{equation}
\begin{array}{r|rrrr}
-3&1 & 3 & -12& 16 \\
& & -3 & 0& 36 \\
\hline & 1 & 0 & -12& 52
\end{array}
\end{equation} The solution is
\begin{equation}
\begin{aligned}
\frac{t^3+3 t^2-12 t+16}{t+3}=t^2-12+\frac{52}{t+3}.
\end{aligned}
\end{equation}
