​ Simplify.

$\quad\frac{\frac{x^2+x-12}{x-2}}{\frac{3x^2+11x-4}{x^2-4}}\\ =\\ \quad\frac{x^2+x-12}{x-2}\cdot\frac{x^2-4}{3x^2+11x-4} \\ =\\ \quad\frac{(x+4)(x-3)}{(x-2)}\cdot\frac{(x+2)(x-2)}{3x^2+12x-x-4} \\ =\\ \quad\frac{(x+4)(x-3)}{(x-2)}\cdot\frac{(x+2)(x-2)}{3x(x+4)-1(x+4)} \\ =\\ \quad\frac{(x+4)(x-3)}{(x-2)}\cdot\frac{(x+2)(x-2)}{(x+4)(3x-1)} \\ =\\ \quad\frac{(x+4)(x-3)(x+2)(x-2)}{(x-2)(x+4)(3x-1)} \\ =\\ \quad\frac{(x-3)(x+2)}{(3x-1)} \qquad\text{and}\qquad x\neq-4\, \qquad x\neq2\\ =\\ \quad\frac{x^2-x-6}{3x-1}$

$\frac{\frac{x^2+x-12}{x-2}}{\frac{3x^2+11x-4}{x^2-4}}$

Simplify the polynomials.

$\frac{\frac{(x+4)(x-3)}{x-2}}{\frac{(3x-1)(x+4)}{(x+2)(x-2)}}$

Multiply the top and bottom fractions by (x+2)(x-2).

$\frac{(x+4)(x-3)(x+2)}{(3x-1)(x+4)}$

Cross out like terms.

$\frac{(x-3)(x+2)}{(3x-1)}$

Multiply the top polynomial out.

$\frac{x^2-x+6}{3x-1}$

Just a different way to look at it. :D