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

$(a-2)(a-3)(a+3)$
Rearranging the terms, the given expression is equivalent to \begin{array}{l}\require{cancel} 18+a^3-9a-2a^2 \\\\= a^3-2a^2-9a+18 .\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} a^3-2a^2-9a+18 \\\\= (a^3-2a^2)-(9a-18) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} a^2(a-2)-9(a-2) .\end{array} Factoring the $GCF= (a-2)$ of the entire expression above results to \begin{array}{l}\require{cancel} (a-2)(a^2-9) .\end{array} The expressions $a^2$ and $9$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $a^2-9 ,$ is a difference of $2$ squares. Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to \begin{array}{l}\require{cancel} (a-2)(a^2-9) \\\\= (a-2)[(a)^2-(3)^2] \\\\= (a-2)(a-3)(a+3) .\end{array}