Partial-Fraction Decomposition: Examples


Print-friendly page

Partial-Fraction Decomposition: Examples (page 3 of 3)

Sections: General techniques, How to handle repeated and irreducible factors, Examples

If the denominator of your rational expression has repeated unfactorable quadratics, then you use linear-factor numerators and follow the pattern that we used for repeated linear factors in the denominator; that is, you'll use fractions with increasing powers of the repeated factors in the denominator.

Set up, but do not solve, the decomposition equality for the following:

[ x^4 + 3x - 2 ] / [ (x^2 + 1)^3 (x - 4)^2 ]

Since x² + 1 is not factorable, I'll have to use numerators with linear factors. Then the decomposition set-up looks like this:

A/(x - 4) + B/(x - 4)^2 + (Cx + D)/(x^2 + 1) + (Ex + F)/(x^2 + 1)^2 + (Gx + H)/(x^2 + 1)^3

Thankfully, I don't have to try to solve this one.

One additional note: Partial-fraction decomposition only works for "proper" fractions. That is, if the denominator's degree is not larger than the numerator's degree (so you have, in effect, an "improper" polynomial fraction), then you first have to use long division to get the "mixed number" form of the rational expression. Then decompose the remaining fractional part.

Decompose the following: Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved

The numerator is of degree 5; the denominator is of degree 3. So first I have to do the long division:

long division

The long division rearranges the rational expression to give me:

Now I can decompose the fractional part. The denominator factors as (x² + 1)(x – 2).

x^2 + (2x^2 + x + 5) / [ (x^2 + 1)(x - 2) ]

The x² + 1 is irreducible, so the decomposition will be of the form:

Multiplying out and solving, I get:

2x² + x + 5 = A(x² + 1) + (Bx + C)(x – 2)
x = 2: 8 + 2 + 5 = A(5) + (2B + C)(0), 15 = 5A, and A = 3
x = 0: 0 + 0 + 5 = 3(1) + (0 + C)(0 – 2),
          5 = 3 – 2C, 2 = –2C, and C = –1
x = 1: 2 + 1 + 5 = 3(1 + 1) + (1B – 1)(1 – 2),
          8 = 6 + (B – 1)(–1) = 6 – B + 1,
          8 = 7 – B, 1 = – B, and B = –1

Then the complete expansion is:

The preferred placement of the "minus" signs, either "inside" the fraction or "in front", may vary from text to text.

<< Previous Top | 1 | 2 | 3 | Return to Index

Cite this article as:

Stapel, Elizabeth. "Partial-Fraction Decomposition: Examples." Purplemath. Available from
http://www.purplemath.com/modules/partfrac3.htm. Accessed

Lessons index

Lessons CD

Purplemath:
  Linking to this site
  Printing pages
  Donating
  School licensing

Reviews of
Internet Sites:
   Free Help
   Practice
   Et Cetera

The "Homework
Guidelines"

Study Skills Survey

Tutoring ($$)

This lesson may be printed out for your personal use.

Copyright © 2006-2008 Elizabeth Stapel \| About \| Terms of Use		Feedback \| Error?