Answer
$\frac{(x-4)(x+3)}{(x-1)(2x+1)}$
Work Step by Step
1. Factor the rational function of the numerator $\frac{2x^2-x-28}{3x^2-x-2}=\frac{(2x+7)(x-4)}{(3x+2)(x-1)}$
2. Factor the rational function of the denominator $\frac{4x^2+16x+7}{3x^2+11x+6}=\frac{(2x+7)(2x+1)}{(3x+2)(x+3)}$
3. Combine the above results $\frac{\frac{2x^2-x-28}{3x^2-x-2}}{\frac{4x^2+16x+7}{3x^2+11x+6}}=\frac{(2x+7)(x-4)}{(3x+2)(x-1)}\times\frac{(3x+2)(x+3)}{(2x+7)(2x+1)}=\frac{(x-4)(x+3)}{(x-1)(2x+1)}$
