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

$(y+4)(2y-1)$
Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} 2y^2+8y-y-4 \\\\= (2y^2+8y)-(y+4) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 2y(y+4)-(y+4) .\end{array} Factoring the $GCF= (y+4)$ of the entire expression above results to \begin{array}{l}\require{cancel} (y+4)(2y-1) .\end{array}