Chapter 2 - Equations, Inequalities, and Problem Solving - 2.1 Solving Equations - 2.1 Exercise Set - Page 87: 91

$5\left( 7a+11c+1 \right)$

The $GCF$ of the given expression, $ 35a+55c+5 ,$ is $5,$ since it is the highest number that can evenly divide (no remainder) all the terms. Factoring the $GCF$ results to \begin{array}{l}\require{cancel} 5\left( \dfrac{35a}{5}+\dfrac{55c}{5}+\dfrac{5}{5} \right) \\\\= 5\left( 7a+11c+1 \right) .\end{array}