Answer
$2(6a-7)(a+1)$
Work Step by Step
Let $z=
2a-3
.$ The given expression, $
3(2a-3)^2+17(2a-3)+10
,$ is equivalent to
\begin{align*}
3z^2+17z+10
.\end{align*}
Using the factoring of trinomials in the form $ax^2+bx+c,$ the expression above has $ac=
3(10)=30
$ and $b=
17
.$
The two numbers with a product of $ac$ and a sum of $b$ are $\left\{
2,15
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{align*}
3z^2+2z+15z+10
.\end{align*}
Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to
\begin{align*}
(3z^2+2z)+(15z+10)
.\end{align*}
Factoring the $GCF$ in each group results to
\begin{align*}
z(3z+2)+5(3z+2)
.\end{align*}
Factoring the $GCF=
(3z+2)
$ of the entire expression above results to
\begin{align*}
(3z+2)(z+5)
.\end{align*}
Substituting back $z=
2a-3
,$ the expression above is equivalent to
\begin{align*}
&
(3(2a-3)+2)(2a-3+5)
\\&=
(6a-9+2)(2a-3+5)
\\&=
(6a-7)(2a+2)
\\&=
2(6a-7)(a+1)
&\text{ (factor $GCF=2$)}
.\end{align*}