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

$(n+5)(7n-1)$
Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} 7n^2+35n-n-5 \\\\= (7n^2+35n)-(n+5) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 7n(n+5)-(n+5) .\end{array} Factoring the $GCF= (n+5)$ of the entire expression above results to \begin{array}{l}\require{cancel} (n+5)(7n-1) .\end{array}