## Intermediate Algebra (12th Edition)

$(3x-5)(2x+7)$
$\bf{\text{Solution Outline:}}$ To factor the given expression, $6x^2+11x-35 ,$ find two numbers whose product is $ac$ and whose sum is $b$ in the quadratic expression $ax^2+bx+c.$ Use these $2$ numbers to decompose the middle term of the given quadratic expression and then use factoring by grouping. $\bf{\text{Solution Details:}}$ Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $6(-35)=-210$ and the value of $b$ is $11 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ -10,21 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 6x^2-10x+21x-35 .\end{array} Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} (6x^2-10x)+(21x-35) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 2x(3x-5)+7(3x-5) .\end{array} Factoring the $GCF= (3x-5)$ of the entire expression above results to \begin{array}{l}\require{cancel} (3x-5)(2x+7) .\end{array}