How do you divide \frac{x^2-25}{x+3} \-: (x-5)?

Dec 25, 2014

We can use the rule about division of rational expressions where you can change the division in a multiplication by flipping the second fraction (where, in your case, the second "fraction" can be written as $\frac{x - 5}{1}$).

In our case you have:

$\frac{{x}^{2} - 25}{x + 3} \div \frac{x - 5}{1} = \frac{{x}^{2} - 25}{x + 3} \times \frac{1}{x - 5}$

We can now manipulate the numerator of the first fraction as:

${x}^{2} - 25 = \left(x + 5\right) \cdot \left(x - 5\right)$

Substituting and simplifying:

$\frac{\left(x + 5\right) \cdot \left(x - 5\right)}{x + 3} \times \frac{1}{x - 5} = \frac{x + 5}{x + 3}$