Answer
$$\frac{\left(2x-1\right)\left(x-2\right)}{3x^2+27x}$$
Work Step by Step
Trying to find like factors in the numerator and the denominator, we obtain:
$$\frac{\left(2x^2-7x+3\right)}{x^2+5x-36}\times \frac{x^2-6x+8}{\left(3x^2-9x\right)} \\ \frac{\left(2x-1\right)\left(x-3\right)\left(x-2\right)\left(x-4\right)}{\left(x-4\right)\left(x+9\right)\times \:3x\left(x-3\right)} \\ \frac{\left(2x-1\right)\left(x-2\right)}{3x^2+27x}$$
