#### Answer

$(2y+1)(3y-1)
$

#### Work Step by Step

Using factoring of trinomials in the form $ax^2+bx+c,$ the given expression, $
6y^2+y-1
,$ has $ac$ equal to $
6(-1)=-6
$ and $b$ equal to $
1
.$
The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{
3,-2
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
6y^2+3y-2y-1
.\end{array}
Grouping the first and second terms and the third and fourth terms, the expression above is equivalent to
\begin{array}{l}\require{cancel}
(6y^2+3y)-(2y+1)
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
3y(2y+1)-(2y+1)
.\end{array}
Factoring the $GCF=
(2y+1)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(2y+1)(3y-1)
.\end{array}