# How do you add or subtract (x^2+3x+2)/(x^2-16)+(3x+6)/(x^2-16)?

May 18, 2015

Since both fractions have the same denominator, you can simply add up (or substract) the numerators and keep the denominator unchanged :

$\frac{{x}^{2} + 3 x + 2}{{x}^{2} - 16} + \frac{3 x + 6}{{x}^{2} - 16} = \frac{\left({x}^{2} + 3 x + 2\right) + \left(3 x + 6\right)}{{x}^{2} - 16}$

$= \frac{{x}^{2} + 6 x + 8}{{x}^{2} - 16}$.

Then, you can factorize it :

$\frac{{x}^{2} + 6 x + 8}{{x}^{2} - 16} = \frac{\left(x + 2\right) \left(x + 4\right)}{\left(x + 4\right) \left(x - 4\right)} = \frac{x + 2}{x - 4}$.

In the case of a substraction :

$\frac{{x}^{2} + 3 x + 2}{{x}^{2} - 16} - \frac{3 x + 6}{{x}^{2} - 16} = \frac{\left({x}^{2} + 3 x + 2\right) - \left(3 x + 6\right)}{{x}^{2} - 16}$

$= \frac{{x}^{2} - 4}{{x}^{2} - 16} = \frac{\left(x + 2\right) \left(x - 2\right)}{\left(x + 4\right) \left(x - 4\right)}$.