## Algebra 2 (1st Edition)

$$\frac{3\left(x-6\right)}{x\left(x+5\right)}$$
Multiplying the two expressions and then cancelling out like terms and factors in the numerator and the denominator, we find: $$\frac{x^2-2x-24}{x^2+3x-10}\cdot \frac{3\left(x-2\right)}{x\left(x+4\right)}\\ \frac{3\left(x+4\right)\left(x-6\right)\left(x-2\right)}{\left(x^2+3x-10\right)x\left(x+4\right)} \\ \frac{3\left(x+4\right)\left(x-6\right)\left(x-2\right)}{\left(x^2+3x-10\right)x\left(x+4\right)}\\ \frac{3\left(x+4\right)\left(x-6\right)\left(x-2\right)}{x\left(x-2\right)\left(x+5\right)\left(x+4\right)}\\ \frac{3\left(x-6\right)}{x\left(x+5\right)}$$