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### Partial Fractions

Posted: June 2nd, 2011, 4:46 pm
$\frac{x+7}{x^2-x-6} = \frac{2}{x-3}-\frac{1}{x+2}$

Each fraction on the right side of the equation is a partial fraction, and together they make up the partial fraction decomposition on the left side.

Steps
1. Factor the Denominator.
2. Express factored form as a sum of denominators using A and B as numerators and set this equal to the original problem.
3. Eliminate the denominators by multiplying each side by the LCD.
4. Find the excluded values of x. (What would make the denominator equal 0?)
5. Substitute the excluded values into the equation and solve for A and B.
6. Substitute the values into step 2.

A. $\frac{x+7}{x^2-x-6} = \frac{A}{x-3}-\frac{B}{x+2}$

$\left [\frac{x+7}{x^2-x-6} = \frac{A}{x-3}-\frac{B}{x+2} \right ](x-3)(x+2)$

$x+7 = A(x-2)+B(x-3)$

x = −2
$-2+7=A(-2+2)+B(-2-3)$

5 = −5B

B = −1

x = 3
$3+7=A(3+2)+B(3-3)$

10 = 5A

A = 2

$\frac{2}{x-3}-\frac{1}{x+2}$

B. $\frac{x+8}{x^2+6x+8} = \frac{A}{x+4}-\frac{B}{x+2}$

$\left [\frac{x+8}{x^2+6x+8} = \frac{A}{x+4}-\frac{B}{x+2} \right ](x+4)(x+2)$

$x+8 = A(x+2)+B(x+4)$

x = −4
$-4+8=A(-4+2)+B(-4+4)$

4 = −2A

A = −2

x = −2
$-2+8=A(-2+2)+B(-2+4)$

6 = 2B

B = 3

$\frac{-2}{x+4}-\frac{3}{x+2}$