Answer
$$\frac{\left(x^2-5\right)\left(x+3\right)}{x^2\left(x^2-25\right)}$$
Work Step by Step
Using the rules of exponents and cancelling common factors in the numerator and the denominator, we find:
$$\frac{\left(2x^2-10\right)\left(x+3\right)}{\left(x^2-25\right)\times \:2x^2} \\ \frac{2\left(x^2-5\right)\left(x+3\right)}{\left(x^2-25\right)\times \:2x^2} \\ \frac{\left(x^2-5\right)\left(x+3\right)}{x^2\left(x^2-25\right)}$$
