## Intermediate Algebra (12th Edition)

$(p+1)(3p-4)$
$\bf{\text{Solution Outline:}}$ To factor the given expression, $3p^2-p-4 ,$ 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. $\bf{\text{Solution Details:}}$ To factor the trinomial expression above, note that the value of $ac$ is $3(-4)=-12$ and the value of $b$ is $-1 .$ 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 $\{ 3,-4 \}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 3p^2+3p-4p-4 .\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} (3p^2+3p)-(4p+4) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 3p(p+1)-4(p+1) .\end{array} Factoring the $GCF= (p+1)$ of the entire expression above results to \begin{array}{l}\require{cancel} (p+1)(3p-4) .\end{array}