# Partial Fraction Decomposition

• Feb 8th 2007, 11:15 AM
spiritualfields
Partial Fraction Decomposition
I need some help understanding an apparent discrepancy. In this problem:

4 / (2x^2 - 5x -3) (2x^2 stands for 2x squared)

I find that if I resolve that rational expression to these partial fractions:

(A / 2x + 1) + (B / x - 3)

I get A = -8/7 and B = 4/7

However, if I resolve the rational expression to these partial fractions:

(A / x -3) + (B / 2x + 1)

I get A = 4/7 and B = -8/7

Then, if I let x = 0 and sum the partial fractions I get -4/3 for the first case, and 20/21 for the second case.

I think the first case is correct because the original rational expression (4 / (2x^2 -5x -3)) does resolve to -4/3 when x is set to 0.

So my question is, how does one determine what order to use when resolving rational expressions into partial fractions composed of linear non-repeating factors? It seems to make a big difference in this example, yet my book does not provide guidance in this area.
• Feb 8th 2007, 11:30 AM
CaptainBlack
Quote:

Originally Posted by spiritualfields
I need some help understanding an apparent discrepancy. In this problem:

4 / (2x^2 - 5x -3) (2x^2 stands for 2x squared)

I find that if I resolve that rational expression to these partial fractions:

(A / 2x + 1) + (B / x - 3)

I get A = -8/7 and B = 4/7

However, if I resolve the rational expression to these partial fractions:

(A / x -3) + (B / 2x + 1)

I get A = 4/7 and B = -8/7

These are the same partial fraction resolutions of the original term.

Quote:

Then, if I let x = 0 and sum the partial fractions I get -4/3 for the first case, and 20/21 for the second case.

I think the first case is correct because the original rational expression (4 / (2x^2 -5x -3)) does resolve to -4/3 when x is set to 0.
Just check your arithmetic, since the two expressions are identical they
must give the same result.

Quote:

So my question is, how does one determine what order to use when resolving rational expressions into partial fractions composed of linear non-repeating factors? It seems to make a big difference in this example, yet my book does not provide guidance in this area.
Either will do, as you see here you get the same result either way.

RonL
• Feb 8th 2007, 11:50 AM
spiritualfields
Yes, I realize now that I had made an arithmetic error. Thanks.