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Topic review (newest first)
- mathsyperson
- 2005-10-25 05:44:32
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. 
- ryos
- 2005-10-25 01:59:15
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.
- Andorin
- 2005-10-25 01:05:27
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?
Cheers in advance