## Intermediate Algebra (12th Edition)

$(9z-1)(3z+5)$
$\bf{\text{Solution Outline:}}$ To factor the quadratic expression $ax^2+bx+c,$ find two numbers whose product is $ac$ and whose sum is $b$. Use these $2$ numbers to decompose the middle term of the quadratic expression and then use factoring by grouping. $\bf{\text{Solution Details:}}$ In the given expression, $27z^2+42z-5 ,$ the value of $ac$ is $27(-5)=-135$ and the value of $b$ is $42 .$ The $2$ numbers that have a product $ac$ and a sum of $b$ are $\{ -3,45 \}.$ Using these $2$ numbers to decompose the middle term of the given expression results to \begin{array}{l}\require{cancel} 27z^2-3z+45z-5 .\end{array} Grouping the first and third terms and the second and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} (27z^2-3z)+(45z-5) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 3z(9z-1)+5(9z-1) .\end{array} Factoring the $GCF= (9z-1)$ of the entire expression above results to \begin{array}{l}\require{cancel} (9z-1)(3z+5) .\end{array}