## College Algebra (10th Edition)

$\displaystyle \frac{(x-4)(x+3)}{(x-1)(2x+1)}$
$\displaystyle \frac{\frac{2x^{2}-x-28}{3x^{2}-x-2}}{\frac{4x^{2}+16x+7}{3x^{2}+11x+6}}=\frac{2x^{2}-x-28}{3x^{2}-x-2}\div\frac{4x^{2}+16x+7}{3x^{2}+11x+6}=$ $=\displaystyle \frac{2x^{2}-x-28}{3x^{2}-x-2}\cdot\frac{3x^{2}+11x+6}{4x^{2}+16x+7}=...$ Factor each trinomial (for $ax^{2}+bx+c$ we search for factors of ac that add up to b.) $2x^{2}-x-28=2x^{2}-8x+7x-28$ $=2x(x-4)+7(x-4)$ $=(2x+7)(x-4)$ $3x^{2}-x-2=3x^{2}-3x+2x-2$ $=3x(x-1)+2(x-1)$ $=(3x+2)(x-1)$ $3x^{2}+11x+6=3x^{2}+9x+2x+6$ $=3x(x+3)+2(x+3)$ $=(3x+2)(x+3)$ $4x^{2}+16x+7=4x^{2}+2x+14x+7$ $=2x(2x+1)+7(2x+1)$ $=(2x+7)(2x+1)$ $...=\displaystyle \frac{(2x+7)(x-4)}{(3x+2)(x-1)}\cdot\frac{(3x+2)(x+3)}{(2x+7)(2x+1)}=$ cancel the common terms $(2x+7)$ and $(3x+2)$ $= \displaystyle \frac{(x-4)(x+3)}{(x-1)(2x+1)}$