# Partial fractions

#### boks

Joined Oct 10, 2008
218
$\frac{s-1}{s(s-2)^2}$

How can I expand this fraction?

$\frac{A}{s} + \frac{B}{(s-2)} + \frac{C}{(s-2)^2}$

right?

This gives me the equation

$As^3 - 6As^2 + 12As - 8A Bs^3 - 4Bs^2 + 4Bs + Cs^2 - 2Cs = s-1$

so that

(1) A + B =0
(2)- 6A - 4B + C = 0
(3) 12A + 4B - 2C = 1
(4) -8A = -1

(4) gives A = 1/8
(1) gives B = -1/8
(2) gives C = 1/4

A = -1/4
B = 1/4
C = 2/4

What's wrong?

Last edited:

#### studiot

Joined Nov 9, 2007
5,003
Yes
I make it

A=-1
B=+1
C=+1

#### vvkannan

Joined Aug 9, 2008
138
I dont think you have got the equation right.
The equation should be
As^2-4As+4A+Bs^2-2Bs+Cs = s-1

#### boks

Joined Oct 10, 2008
218
Thanks, I figured it out now.

#### studiot

Joined Nov 9, 2007
5,003
OOps

What sloppy arithmetic Studio T

Go to the bottom of the class.

Boks original was correct.