# How do you write the partial fraction decomposition of the rational expression  (x+10)/(x^2+2x-8)?

Dec 9, 2015

The equivalent partial fraction is :
$\frac{- 1}{x + 4} + \frac{2}{x - 2}$

#### Explanation:

Given $\frac{x + 10}{{x}^{2} + 2 x - 8}$

Step 1: Factor the denominator

(x+10)/((x+4)(x-2)

Step 2: Set up the partial faction as follows:

$\frac{x + 10}{\left(x + 4\right) \left(x - 2\right)} = \frac{A}{x + 4} + \frac{B}{x - 2} \text{ " " " } \left(1\right)$

Step 3: Multiply both sides by the LCD, $\left(x + 4\right) \left(x - 2\right)$:

$\left(x + 10\right) = A \left(x - 2\right) + B \left(x + 4\right)$
$x + 10 = A x - 2 A + B x + 4 B$

Step 4: Set up a system like this
$1 x : \text{ " " " A+ B= 1 " " " } \left(2\right)$
$10 : \text{ " " -2A+4B= 10 " " " } \left(3\right)$

Step 5. You can solve the system by the elimination method:

$2 \left(A + B = 1\right) \implies 2 A + 2 B = 2$

$+ - 2 A + 4 B = 10$
$6 B = 12 \implies B = 2$

Solve for $A$ by substituting $B = 3$ into $\left(2\right)$:

$A + \left(2\right) = 1$
$A = - 1$

Step 6. Substitute $A$ and $B$ back into $\left(1\right)$:

$\frac{x + 10}{\left(x + 4\right) \left(x - 2\right)} = \frac{- 1}{x + 4} + \frac{2}{x - 2}$