Answer
$$\left(x+2\right)\left(x+7\right)$$
Work Step by Step
Using the rules of exponents and cancelling common factors in the numerator and the denominator, we find:
$$\frac{x^2-3x-10}{x^2-2x-15}\left(x^2+10x+21\right)\\ \frac{x+2}{x+3}\left(x^2+10x+21\right)\\ \frac{\left(x+2\right)\left(x+3\right)\left(x+7\right)}{x+3}\\ \left(x+2\right)\left(x+7\right)$$
