# Partial Fractions

• Feb 4th 2010, 03:13 AM
Paymemoney
Partial Fractions
Hi
Need help on expressing the following equation in partial fractions:
$\frac{x^2-x+3}{(x^2-3x+7)(x+2)^2)}$
$=\frac{A}{(x+2)}+\frac{B}{(x+2)^2}+\frac{Cx+D}{x^2-3x+7}$

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

Equation 1: $A+C=0$
Equation 2: $-A+B+4C+D=1$
Equation 3: $A-3B+4C+4D=-1$
Equation 4: $14A+7B+4D=3$

Equation 2-3
$-A+B+4C+D=1$- $A-3B+4C+4D=-1$
Gets: $-2A+4B-3D=2$

Equation (2-3)-4
$-2A+4B-3D=2$- $14A+7B+4D=3$
$35B-17D=17$

Don't know what to do next??

P.S
• Feb 4th 2010, 04:25 AM
HallsofIvy
Quote:

Originally Posted by Paymemoney
Hi
Need help on expressing the following equation in partial fractions:
$\frac{x^2-x+3}{(x^2-3x+7)(x+2)^2)}$
$=\frac{A}{(x+2)}+\frac{B}{(x+2)^2}+\frac{Cx+D}{x^2-3x+7}$

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

Equation 1: $A+C=0$
Equation 2: $-A+B+4C+D=1$
Equation 3: $A-3B+4C+4D=-1$
Equation 4: $14A+7B+4D=3$

Equation 2-3
$-A+B+4C+D=1$- $A-3B+4C+4D=-1$
Gets: $-2A+4B-3D=2$

Equation (2-3)-4
$-2A+4B-3D=2$- $14A+7B+4D=3$
$35B-17D=17$

Don't know what to do next??

P.S

You have eliminated A and C from your equation but have not yet used equation 1. Use that (which involves only A and C) together with the other equations to get a second equation in B and D only. Then combine the two equations to eliminate either B or D so you can solve for the remaining value.
• Feb 4th 2010, 07:24 PM
Paymemoney
ok thanks, i'll try that