Answer
$$\frac{2x\left(x+4\right)}{\left(x+2\right)\left(x-3\right)}$$
Work Step by Step
Using the rules of exponents and cancelling common factors in the numerator and the denominator, we find:
$$\frac{\left(x^2+3x-4\right)\left(2x^2+4x\right)}{\left(x^2+4x+4\right)\left(x^2-4x+3\right)}\\ \frac{2x\left(x-1\right)\left(x+4\right)\left(x+2\right)}{\left(x+2\right)^2\left(x-1\right)\left(x-3\right)}\\ \frac{2x\left(x+4\right)}{\left(x+2\right)\left(x-3\right)}$$
