## Intermediate Algebra (12th Edition)

$6z(2z^2-z+3)$
$\bf{\text{Solution Outline:}}$ To factor the given expression, $12z^3-6z^2+18z ,$ factor first the $GCF.$ Then use the factoring of trinomials in the form $ax^2+bx+c.$ $\bf{\text{Solution Details:}}$ The $GCF$ of the terms in the given expression is $6z ,$ since it is the greatest expression that can divide all the terms evenly (no remainder.) Factoring the $GCF$ results to \begin{array}{l}\require{cancel} 6z(2z^2-z+3) .\end{array} In the trinomial expression above, $a= 2 ,b= -1 ,\text{ and } c= 3 .$ There are no two numbers whose product is $ac= 2(3)=6$ and whose sum is $b.$ Hence, the factored form is \begin{array}{l}\require{cancel} 6z(2z^2-z+3) .\end{array}