## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$3x(2x+3)(3x-1)$
Factoring the $GCF= 3x ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 18x^3+21x^2-9x \\\\= 3x(6x^2+7x-3) .\end{array} Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} 3x(6x^2+7x-3) \end{array} has $ac= 6(-3)=-18$ and $b= 7 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ 9,-2 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 3x(6x^2+9x-2x-3) .\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} 3x[(6x^2+9x)-(2x+3)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 3x[3x(2x+3)-(2x+3)] .\end{array} Factoring the $GCF= (2x+3)$ of the entire expression above results to \begin{array}{l}\require{cancel} 3x[(2x+3)(3x-1)] \\\\= 3x(2x+3)(3x-1) .\end{array}