## Intermediate Algebra (12th Edition)

$(3r+16)(r+1)$
$\bf{\text{Solution Outline:}}$ To factor the given expression, $3(r+5)^2-11(r+5)-4 ,$ use substitution to simplify the expression. Then use factoring of trinomials by finding 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. The use back-substitution to have the factored form of the original expression. $\bf{\text{Solution Details:}}$ Let $z=(r+5).$ Then the given expression is equivalent to \begin{array}{l}\require{cancel} 3z^2-11z-4 .\end{array} To factor the trinomial expression above, note that the value of $ac$ is $3(-4)=-12$ and the value of $b$ is $-11 .$ The possible pairs of integers whose product is $ac$ are \begin{array}{l}\require{cancel} \{1,-12\}, \{2,-6\}, \{3,-4\}, \\ \{-1,12\}, \{-2,6\}, \{-3,4\} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ 1,-12 \}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 3z^2+z-12z-4 .\end{array} Grouping the first and second terms and the third and fourth terms of the trinomial, the given expression is equivalent to \begin{array}{l}\require{cancel} (3z^2+z)-(12z+4) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} z(3z+1)-4(3z+1) .\end{array} Factoring the $GCF= (3z+1)$ of the entire expression above results to \begin{array}{l}\require{cancel} (3z+1)(z-4) .\end{array} Since $z=r+5,$ then by back-substitution, the expression above is equivalent to \begin{array}{l}\require{cancel} (3(r+5)+1)((r+5)-4) \\\\= (3r+15+1)(r+5-4) \\\\= (3r+16)(r+1) .\end{array}