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## #1 2005-10-25 01:05:27

Andorin
Novice

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### Partial Fractions with Repeated Factors

Hi,

I'm trying to solve a particular partial fraction namely:

2X^2 - X + 1
-------------------
(X + 1)(X - 1)^2

I am aware of the fractions that it is meant to split into... the numerators are all meant to be 1.

Its how on earth they get there that confuses me.

I can work C out (in terms of there being three constants/numerators named A, B and C) as being 1 but then I get A as being 1/2... what am I doing wrong??? Or is the book I'm working from barkers?

Could someone run me through this one?

## #2 2005-10-25 01:59:15

ryos
Power Member

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### Re: Partial Fractions with Repeated Factors

This one doesn't need A's and C's. You can just break it up, like so:

2x²                   x                       1
-------------  -  --------------  +  ---------------
(x+1)(x-1)²     (x+1)(x-1)²       (x+1)(x-1)²

Expand the denominator:

2x²                        x                         1
------------------  -  -------------------  +  ---------------
x³ - 3x² - x + 1      x³ - 3x² - x + 1       (x+1)(x-1)²

2x²
------------------------------------------
2x²[ (x/2) - (3/2) - (1/2x) + (1/2x²) ]

x
--------------------------
x[ x² - 3x + (1/x) - 1 ]

1                                       1                            1
------------------------------------  -  ----------------------  +  ---------------
(x/2) - (3/2) - (1/2x) + (1/2x²)      x² - 3x + (1/x) - 1        (x+1)(x-1)²

Ugly but true.

Last edited by ryos (2005-10-25 02:05:17)

El que pega primero pega dos veces.

## #3 2005-10-25 05:44:32

mathsyperson
Moderator

Offline

### Re: Partial Fractions with Repeated Factors

I think that's far more complicated than it needs to be.

2X² - X + 1
-------------------
(X+1)(X - 1)²

Split up into partial fractions:      A           B           C
-----  +  -----  +  -------
X+1       X-1       (X-1)²

Multiply out:  A(X-1)² + B(X+1)(X-1) + C(X+1) = 2X² - X + 1

Substitute in X = 1 : 2C = 2   ∴  C = 1
Substitute in X = -1: 4A = 4   ∴  A = 1
Find B: X² - 2X + 1 + B(X² - 1) + X + 1 = 2X² - X + 1
Simplify: B(X² - 1) = X² - 1  ∴  B = 1

Therefore,   2X² - X + 1             1           1           1
-----------------        -----  +  -----  +  -------
( X+1)(X-1)²          X+1       X-1       (X-1)²

Not as ugly, still true.

Why did the vector cross the road?
It wanted to be normal.