## Intermediate Algebra (12th Edition)

$(b+3)(x+y)$
$\bf{\text{Solution Outline:}}$ Group the terms of the given expression, $3x+by+bx+3y ,$ such that the factored form of the groupings will result to a factor that is common to the entire expression. Then, factor the $GCF$ in each group. Finally, factor the $GCF$ of the entire expression. $\bf{\text{Solution Details:}}$ Grouping the first and third terms and the second and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} (3x+bx)+(by+3y) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} x(3+b)+y(b+3) \\\\= x(b+3)+y(b+3) .\end{array} Factoring the $GCF= (b+3)$ of the entire expression above results to \begin{array}{l}\require{cancel} (b+3)(x+y) .\end{array}