## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

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