Answer
$5(t+3)(3t-5)$
Work Step by Step
Factoring the $GCF=
5
,$ the given expression is equivalent to
\begin{array}{l}\require{cancel}
15t^2+20t-75
\\\\=
5(3t^2+4t-15)
.\end{array}
Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{
expression
}$
\begin{array}{l}\require{cancel}
5(3t^2+4t-15)
\end{array} has $ac=
3(-15)=-45
$ and $b=
4
.$
The two numbers with a product of $c$ and a sum of $b$ are $\left\{
9,-5
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
5(3t^2+9t-5t-15)
.\end{array}
Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to
\begin{array}{l}\require{cancel}
5[(3t^2+9t)-(5t+15)]
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
5[3t(t+3)-5(t+3)]
.\end{array}
Factoring the $GCF=
(t+2)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
5[(t+3)(3t-5)]
\\\\=
5(t+3)(3t-5)
.\end{array}