#### Answer

$3(9r-10)(2r+1)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To factor the given expression, $
6(3r-1)^2+(3r-1)-35
,$ use substitution to make the given expression a trinomial. Then, find two numbers whose product is $ac$ and whose sum is $b$ in the quadratic expression $ax^2+bx+c.$ Use these $2$ numbers to decompose the middle term of the given quadratic expression and then use factoring by grouping. Finally, use back-substitution to get the factored form of the original expression.
$\bf{\text{Solution Details:}}$
Let $z=(3r-1).$ Then, the expression above is equivalent to
\begin{array}{l}\require{cancel}
6z^2+z-35
.\end{array}
In the trinomial expression above the value of $ac$ is $
6(-35)=210
$ and the value of $b$ is $
1
.$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\{
-14,15
\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
6z^2-14z+15z-35
.\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}
(6z^2-14z)+(15z-35)
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
2z(3z-7)+5(3z-7)
.\end{array}
Factoring the $GCF=
(3z-7)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(3z-7)(2z+5)
.\end{array}
Since $z=(3r-1)$, then the expression above is equivalent to
\begin{array}{l}\require{cancel}
[3(3r-1)-7][2(3r-1)+5]
\\\\=
[9r-3-7][6r-2+5]
\\\\=
[9r-10][6r+3]
\\\\=
[9r-10]3[2r+1]
\\\\=
3(9r-10)(2r+1)
.\end{array}