## Precalculus: Mathematics for Calculus, 7th Edition

$(w-2)(w+2)(w-3)$
$w^{3}-3w^{2}-4w+12$ Factor this expression by grouping. Begin by grouping the first two terms and the last two terms together: $w^{3}-3w^{2}-4w+12=(w^{3}-3w^{2})-(4w-12)=...$ Take out common factor $w^{2}$ from the first parentheses and $4$ from the second parentheses: $...=w^{2}(w-3)-4(w-3)=...$ Take out common factor $w-3$: $...=(w^{2}-4)(w-3)=...$ Factor the difference of squares in the first parentheses. The formula for factoring an expression like this is $A^{2}-B^{2}=(A-B)(A+B)$. For the expression $A^{2}=w^{2}$ and $B^{2}=4$. $...=(w-2)(w+2)(w-3)$