Answer
$(t+4)(5t+8)$
Work Step by Step
Using the factoring of trinomials in the form $ax^2+bx+c,$ the given expression,
\begin{align*}
5t^2+28t+32
\end{align*}
has $ac=
5(32)=160
$ and $b=
28
.$
The two numbers with a product of $c$ and a sum of $b$ are $\left\{
8,20
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{align*}
5t^2+20t+8t+32
\end{align*}
Grouping the first and second terms and the third and fourth terms, the expression above is equivalent to
\begin{align*}
(5t^2+20t)+(8t+32)
\end{align*}
Factoring the $GCF$ in each group results to
\begin{align*}
5t(t+4)+8(t+4)
\end{align*}
Factoring the $GCF=
(t+4)
$ of the entire expression above results to
\begin{align*}
(t+4)(5t+8)
\end{align*}
