## Introductory Algebra for College Students (7th Edition)

$(4y-1)(3y-2)$
$12y^{2}-11y+2$ Two factors of $ac=24$, whose sum is $b=-11$ are $-3$ and $-8$. Rewrite the middle term and factor in pairs. $12y^{2}-11y+2$ $=12y^{2}-3y-8y+2$ $=3y(4y-1)-2(4y-1)]\qquad$...factor out the common binomial factor,$4y-1$. $=(4y-1)(3y-2)$