# Partial fractions - having 1 in the numerator?

1. Oct 7, 2008

### paul34

This is probably a "basic" question, but I can't seem to remember how to do partial fractions problems where there is only a 1 in the numerator.

For example (just making this up), let's say I have:

1/s(s+4)(s+5)

So what I'd do is 1/s(s+4)(s+5) = A/s + B/(s+4) + C/(s+5) as one would expect, but this can't be right, correct? It seems a lot of coefficients don't "work out."

For example, after doing cross multiplications and combining terms, you would basically get:

For s^2 terms: (A+B+C) = 0 [1]
For s terms: (6A + 5B + 4C) = 0 [2]
For no-s terms: 5A = 1 [3]

Then putting [3]--> [1], then getting B=-C - 1/5, and putting that in [2]. You'll end up with A being equal to 1/5, and C being equal to -1/5. Plug this all into [1], so then that means B is equal to zero. What am I missing here? Is this right?

Matlab says it should be C=-1 and B=1, which means that A is zero then? Huh? I am pretty confused.

Thanks.

Last edited: Oct 7, 2008
2. Oct 7, 2008

### Staff: Mentor

You have decomposed the expression correctly, but have made a mistake in your work. Not only that, but the results you reported from Matlab are wrong, too.
Your equation [1] is correct, equations [2] and [3] are wrong. The coefficient for the s terms is 9A + 5B + 4C, and for the constant term, the coefficient is 20.

So the three equations are:
Code (Text):

A + B + C = 0
9A + 5B + 4C = 0
20A = 1

From these, you should get A = 1/20, B = -1/4, and C = 1/5
Mark

3. Oct 8, 2008

Thanks Mark.