# How do you combine (3n + 8)/(n^2 + 6n + 8) - ( 4n + 2)/(n^2 + n - 12)?

Apr 17, 2018

$\frac{3 n + 8}{{n}^{2} + 6 n + 8} - \frac{4 n - 2}{{n}^{2} + n - 12} = \frac{{n}^{2} + 7 n + 20}{\left(n + 4\right) \left(n + 2\right) \left(3 - n\right)}$

#### Explanation:

$\frac{3 n + 8}{{n}^{2} + 6 n + 8} - \frac{4 n - 2}{{n}^{2} + n - 12}$

First factor the denominators.

$\frac{3 n + 8}{\left(n + 4\right) \left(n + 2\right)} - \frac{4 n - 2}{\left(n + 4\right) \left(n - 3\right)}$

The common denominator is $\left(n + 4\right) \left(n + 2\right) \left(n - 3\right)$

Multiply each quotient by the appropriate factor to display the common denominator

$\frac{\left(3 n + 8\right)}{\left(n + 4\right) \left(n + 2\right)} \frac{\left(n - 3\right)}{\left(n - 3\right)} - \frac{\left(4 n - 2\right)}{\left(n + 4\right) \left(n - 3\right)} \frac{\left(n + 2\right)}{\left(n + 2\right)}$

Expand the numerators using the distributive property (or FOIL if you like).

$\frac{3 {n}^{2} - n - 24}{\left(n + 4\right) \left(n + 2\right) \left(n - 3\right)} - \frac{4 {n}^{2} + 6 n - 4}{\left(n + 4\right) \left(n + 2\right) \left(n - 3\right)}$

We can combine the quotients because they have a common denominator.

$\frac{3 {n}^{2} - n - 24 - 4 {n}^{2} - 6 n + 4}{\left(n + 4\right) \left(n + 2\right) \left(n - 3\right)}$

Combine like terms.

$\frac{- {n}^{2} - 7 n - 20}{\left(n + 4\right) \left(n + 2\right) \left(n - 3\right)} = \frac{{n}^{2} + 7 n + 20}{\left(n + 4\right) \left(n + 2\right) \left(3 - n\right)}$