Answer
$3(2x+5)(x+1)$
Work Step by Step
Factoring the $GCF=
3
,$ the given expression is equivalent to
\begin{array}{l}\require{cancel}
6x^2+21x+15
\\\\=
3(2x^2+7x+5)
.\end{array}
Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{
expression
}$
\begin{array}{l}\require{cancel}
3(2x^2+7x+5)
\end{array} has $ac=
2(5)=10
$ and $b=
7
.$
The two numbers with a product of $c$ and a sum of $b$ are $\left\{
5,2
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
3(2x^2+5x+2x+5)
.\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}
3[(2x^2+5x)+(2x+5)]
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
3[x(2x+5)+(2x+5)]
.\end{array}
Factoring the $GCF=
(x+6)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
3[(2x+5)(x+1)]
\\\\=
3(2x+5)(x+1)
.\end{array}
