# Partial fractions

1. Dec 11, 2008

### leopard

$$\frac{s-1}{s(s-2)^2}$$

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?

2. Dec 11, 2008

### gabbagabbahey

So far so good ...

3. Dec 11, 2008

### leopard

I figured it out, thanks!

4. Dec 11, 2008

### HallsofIvy

Staff Emeritus
leopard, for future use, you may find it easier (and less error prone) to do this:
to find A, B, C so that
$$\frac{A}{s}+ \frac{B}{s-2}+ \frac{C}{(x-2)^2}= \frac{s-1}{s(s-2)^2}$$
Multiply both sides by the denominator, s(s-2)2, leaving it as
$$A(s-2)^2+ Bs(s-2)+ Cs= s- 1$$

Now, since this must be true for all s, select simple values of s:
if s= 0, 4A= -1
if s= 2, 2C= 1
if s= 1, A- B+ C= 0