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

$(w+2)(w-2)(w+3)$
Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} w^3-4w+3w^2-12 \\\\= (w^3-4w)+(3w^2-12) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} w(w^2-4)+3(w^2-4) .\end{array} Factoring the $GCF= (w^2-4)$ of the entire expression above results to \begin{array}{l}\require{cancel} (w^2-4)(w+3) .\end{array} The expressions $w^2$ and $4$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $w^2-4 ,$ 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} [(w)^2-(2)^2](w+3) \\\\= [(w+2)(w-2)](w+3) \\\\= (w+2)(w-2)(w+3) .\end{array}