Answer
$5(a+3)(2a-1)$
Work Step by Step
Factoring the $GCF=
5
,$ the given expression is equivalent to
\begin{array}{l}\require{cancel}
10a^2+25a-15
\\\\=
5(2a^2+5a-3)
.\end{array}
Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{
expression
}$
\begin{array}{l}\require{cancel}
5(2a^2+5a-3)
\end{array} has $ac=
2(-3)=-6
$ and $b=
5
.$
The two numbers with a product of $c$ and a sum of $b$ are $\left\{
6,-1
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
5(2a^2+6a-1a-3)
.\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[(2a^2+6a)-(a+3)]
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
5[2a(a+3)-(a+3)]
.\end{array}
Factoring the $GCF=
(a+3)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
5[(a+3)(2a-1)]
\\\\=
5(a+3)(2a-1)
.\end{array}
