Answer
$ \displaystyle \frac{(x-4)(x+3)}{(x-1)(2x+1)}$
Work Step by Step
$\displaystyle \frac{\frac{2x^{2}-x-28}{3x^{2}-x-2}}{\frac{4x^{2}+16x+7}{3x^{2}+11x+6}}=\frac{2x^{2}-x-28}{3x^{2}-x-2}\div\frac{4x^{2}+16x+7}{3x^{2}+11x+6}=$
$=\displaystyle \frac{2x^{2}-x-28}{3x^{2}-x-2}\cdot\frac{3x^{2}+11x+6}{4x^{2}+16x+7}=...$
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 $(3x+2)$
$= \displaystyle \frac{(x-4)(x+3)}{(x-1)(2x+1)}$
