## College Algebra (11th Edition)

$3(9r-10)(2r+1)$
$\bf{\text{Solution Outline:}}$ To factor the given expression, $6(3r-1)^2+(3r-1)-35 ,$ use substitution to make the given expression a trinomial. Then, 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. Finally, use back-substitution to get the factored form of the original expression. $\bf{\text{Solution Details:}}$ Let $z=(3r-1).$ Then, the expression above is equivalent to \begin{array}{l}\require{cancel} 6z^2+z-35 .\end{array} In the trinomial expression above the value of $ac$ is $6(-35)=210$ and the value of $b$ is $1 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\{ -14,15 \}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 6z^2-14z+15z-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} (6z^2-14z)+(15z-35) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 2z(3z-7)+5(3z-7) .\end{array} Factoring the $GCF= (3z-7)$ of the entire expression above results to \begin{array}{l}\require{cancel} (3z-7)(2z+5) .\end{array} Since $z=(3r-1)$, then the expression above is equivalent to \begin{array}{l}\require{cancel} [3(3r-1)-7][2(3r-1)+5] \\\\= [9r-3-7][6r-2+5] \\\\= [9r-10][6r+3] \\\\= [9r-10]3[2r+1] \\\\= 3(9r-10)(2r+1) .\end{array}