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Partial Fractions

Date: 11/23/2003 at 16:09:41
From: jamie
Subject: Partial Fractions

I need to put this fraction into power series formation:

       3x^2 - x
  x^3 - x^2 - x + 1

I'm not sure how to set up partial fractions because it doesn't 
factor.  The farthest I get on factoring is this:

       3x^2 - x
   x(x^2 - x - 1) + 1

Date: 11/23/2003 at 22:33:00
From: Doctor Jordan
Subject: Re: Partial Fractions

Dear Jamie,

Your thinking about wanting to factor is correct, and you are getting
stuck because you have not been able to factor the denominator, which
can actually be factored.  Let's look at how.

By observation we see that x^3 - x^2 - x + 1 has a value of zero for 
x = 1.  So we know that (x - 1) is a factor of it.  Using long or
synthetic division with (x - 1), we find that it factors into:

  (x - 1)(x^2 - 1)

then noting that the second piece is a difference of squares we get

  (x - 1)(x + 1)(x - 1)

which can be written

  (x - 1)^2(x + 1)

So we set up our partial fractions like this:

     3x^2-x        A        B       C
  -----------  = ------ + ----- + -----
  (x-1)^2(x+1)  (x-1)^2   (x-1)   (x+1)

We see with the A and B fractions that if we have a factor that is 
squared, we must have a variable over each of its decreasing powers, 
from its original power through to 1 (that is, we have A over the 
factor to the power of 2 [its original power], and B over the factor 
to the power of 1).

We can now solve this problem; we multiply A by (x+1), B by (x-1)
(x+1), and C by (x-1)^2.  Doing this achieves a common denominator on 
the right hand side of (x-1)^2(x+1), and since we have the same 
denominator on the left and right sides, we can multiply both sides 
by that denominator to eliminate it.

Then we proceed to substitute in x values, solving this equation the 
standard way for partial fractions.

If you have any other questions, or didn't quite like this 
explanation, please write me back.  Good luck!

- Doctor Jordan, The Math Forum 
Associated Topics:
High School Calculus
High School Polynomials

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