## Precalculus: Mathematics for Calculus, 7th Edition

$3x^{3}-x^{2}-12x+4=(3x-1)(x-2)(x+2)$
$3x^{3}-x^{2}-12x+4$ Group the first and third terms together. Do the same with the second and fourth terms: $3x^{3}-x^{2}-12x+4=(3x^{3}-12x)+(-x^{2}+4)=...$ Take out common factor $-1$ from the second parentheses: $...=(3x^{3}-12x)-(x^{2}-4)=...$ Take out common factor $3x$ from the first parentheses: $...=3x(x^{2}-4)-(x^{2}-4)=...$ Take out common factor $(x^{2}-4)$: $...=(3x-1)(x^{2}-4)=...$ Finally, factor the difference of squares in the second parentheses: $...=(3x-1)(x-2)(x+2)$