Answer
a) $\frac{3t^4-17t^3+47t^2-153t+180}{t^2+9}= 3t^2-17t+20$
b) $ (t^2+9)\cdot (t-4)\cdot(3t-5)$
Work Step by Step
Given \begin{equation}
\left(3t^4-17t^3+47t^2-153t+180\right) \div\left(t^2+9\right).
\end{equation} a) Use long division.
$$
\begin{array}{r}
3t^2-17t+20\phantom{)} \\
t^2+9{\overline{\smash{\big)}\,3t^4-17t^3+47t^2-153t+180\phantom{)}}}\\
\underline{-~\phantom{(}(3t^4+27t^2)\phantom{-b)}}\\
0-17t^3+20t^2-153t\phantom{)}\\
\underline{-~\phantom{()}(-17t^3-153t)}\\
0+20t^2+180\phantom{)}\\
\underline{-~\phantom{()}(20t^2+180)}\\
0\phantom{)}
\end{array}
$$ The solution is \begin{equation}
\frac{3t^4-17t^3+47t^2-153t+180}{t^2+9}= 3t^2-17t+20.
\end{equation} b) Factor the quadratic $3t^2-17t+20$.
\begin{equation}
\begin{aligned}
3t^2-17t+20&=3t^2-12t-5t+20 \\
&=3t(t-4)-5(t-4)\\
&= (t-4)(3t-5).
\end{aligned}
\end{equation} The prime factor decomposition is $$ (t^2+9)\cdot (t-4)\cdot(3t-5).$$
