# Partial fraction decomposition

1. Aug 30, 2005

### TonyC

Having trouble with one more problem:

find the parial fraction decomposition for th erational expression:
-8x^3-2
x^2(x-1)^3

2. Aug 30, 2005

### whozum

$$\frac{-8x^3-2}{x^2(x-1)^3} = \frac{-8x^3-2}{(x+0)^2(x-1)^3}$$

$$\frac{A}{x} + \frac{Bx+C}{x^2} + \frac{D}{x-1}+\frac{Ex+F}{(x-1)^2} + \frac{Gx^2+Hx+I}{(x-1)^3}$$ if I remember correctly.

From there its just like any other PFD problem.

3. Aug 30, 2005

### TonyC

Got it...whiz bang!

4. Aug 30, 2005

### TonyC

I came up with -6/x -2/x^2 + 6/x-1 + 4/(x-1)^2 + 10/(x-1)^3

Well?

5. Aug 30, 2005

### TD

You don't need lineair and quadratic expressions in the nominators.
When the denominator is either lineair (ax+b), or lineair to a certain power (ax+b)^n, the the proposed nominator is just a number.
It's only when the denominator is a quadratic with discriminant < 0 that the nominators is proposed to be lineair. Quadratics in the nominator never appear.

So the proposed partial fractions are:

$$\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-1}+\frac{D}{(x-1)^2} + \frac{E}{(x-1)^3}$$

@Tony: I see a lot of right nominators but many signs are wrong.

6. Aug 30, 2005

### TonyC

I'm still a bit foggy:

$$\frac{A}{x} + \frac{B}{x^2} - \frac{C}{x-1}-\frac{D}{(x-1)^2} - \frac{E}{(x-1)^3}$$

I am coming up with negative numbers each time I run it.

Last edited by a moderator: Aug 30, 2005
7. Aug 30, 2005

### TD

Have you tried with my proposal of partial fractions?

8. Aug 30, 2005

### TonyC

This is what I came up with: I am getting negative signs each time I run it.

$$\frac{A}{x} + \frac{B}{x^2} - \frac{C}{x-1}-\frac{D}{(x-1)^2} - \frac{E}{(x-1)^3}$$

9. Aug 30, 2005

### TD

What do you mean you run? Negatives? You still have the undetermined coeffecients :S

10. Aug 30, 2005

### TonyC

I have the nominators, they aren't a problem. I have run this several times and have come up with different answers each time.
When I said negative, I meant to say minus.

11. Aug 30, 2005

### TD

Well, if you factor again and compare to the initial fraction, you get

$$\frac{{x^4 \left( {a + c} \right) + x^3 \left( { - 3a + b - 2c + d} \right) + x^2 \left( {3a - 3b + c - d + e} \right) + x\left( {3b - a} \right) - b}} {{x^2 \left( {x - 1} \right)^3 }} = \frac{{ - 8x^3 - 2}} {{x^2 \left( {x - 1} \right)^3 }}$$

This can give you a (not too hard) 5x5 system, or you could simplify your calculations a bit by choosing smart x's, i.e. zero's of the denominator.

12. Aug 30, 2005

### TonyC

I am still lost!

13. Aug 30, 2005

### TD

"I came up with -6/x -2/x^2 + 6/x-1 + 4/(x-1)^2 + 10/(x-1)^3

Well?"

14. Aug 30, 2005

### TonyC

After simplifying the equation, I came up with 2 different answers:
This is why I asked for assistance. I am truly lost at this point.

15. Aug 30, 2005

### TD

But what exactly is the problem? How did you manage to get that first answer?
You can probably use the same method to solve what I wrote.

16. Aug 30, 2005

### TonyC

$$\frac{6}{x} + \frac{2}{x^2} - \frac{6}{x-1}-\frac{4}{(x-1)^2} - \frac{10}{(x-1)^3}$$

17. Aug 30, 2005

### HallsofIvy

Staff Emeritus
You have posted a number of times that you get "negative answers" (which isn't necessarily wrong), that you get two different answers, and that you are lost but you still haven't shown what you have done so we could point out possible mistakes!

You want to find A, B, C, D, E so that
$$\frac{-8x^3- 2}{x^2(x-1)^3}= \frac{A}{x}+ \frac{B}{x^2}+ \frac{C}{x-1}+ \frac{D}{(x-1)^2}+ \frac{E}{(x-1)^3}$$

TD wrote the left side of that as a single fraction but I think it is easiest to multiply both sides of that by x2(x-1)3 to get
$$-8x^3- 2$$= Ax(x-1)^3+ B(x-1)^3+ Cx^2(x-1)^2+ Dx^2(x-1)+ Ex^2[/itex]

The easy thing to do is take x= 0 so that
$$-2= -B$$ or B= 2
and then take x= 1 so that
$$-10= E$$.

Since those are the only two x values that make things on the right side 0, you now have two different ways to find A, C, D.
(a) Multiply the whole thing out and combine like powers of x
(b) Use simple values for x, like -1, 2, -2, to get three equation for A, C, D.