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Old 28th November 2011, 12:01 AM   #1 (permalink)
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Partial Fraction of Proper Fraction's factorisation

Hi, I am not sure of a certain question and would like someone to explain a few part of the solution to me, here is the question and the solution:

Express 5x^2 - 8x - 5 / (x - 1)(x^2 - 1) in partial fractions

Here is the solution:
Factorizing the denominator first:

(x - 1)(x^2 - 1) = (x - 1)(x - 1)(x + 1) = (x + 1)(x - 1)^2
Let 5x^2 - 8x - 5 / (x - 1)(x^2 - 1) = A / x + 1 + B / X - 1 + C / (X - 1)^2

Multiply both sides by (X + 1)(X - 1)^2 to remove the denominators of both sides
5x^2 - 8x - 5 = A(x - 1)^2 + B(x + 1)(x - 1) + C(x + 1) ...............(1)
Put X = 1 into (1)
5(1)^2 - 8(1) - 5 = A(1 - 1)^2 + B(1 + 1)(1 - 1) + C(1 + 1)
C = -4 --------------------(2)
and so on....

my question is the one in bold and the factorization,
(x - 1)(x^2 - 1) = (x - 1)(x - 1)(x + 1) = (x + 1)(x - 1)^2
A / x + 1 + B / X - 1 + C / (X - 1)^2

I am not sure why is there another x - 1 since I have workout:
(x^2 - 1) = (x - 1)(x + 1)
therefore (x - 1)(x^2 -1) which is the original denominator in that question and when I factorize out, the answer is (x - 1)(x - 1)(x + 1) which is the same as (x + 1)(x - 1)^2

for the partial fraction rules, I have to write it as
5x^2 - 8x - 5 / (x - 1)(x^2 - 1) = A / x + 1 + B / x - 1 + C / (x - 1)^2
so I am not sure where does this B / x -1 in the denominator comes from? can anyone explain to me?

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Old 28th November 2011, 12:08 AM   #2 (permalink)
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Re: Partial Fraction of Proper Fraction's factorisation

If this is additional mathematics in secondary school, I guess you have to check the textbook. Because it is a law, a must to do so.

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Old 28th November 2011, 01:38 AM   #3 (permalink)
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Re: Partial Fraction of Proper Fraction's factorisation

You have to follow laws of partial faction.
remember that...


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Old 28th November 2011, 01:42 AM   #4 (permalink)
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Re: Partial Fraction of Proper Fraction's factorisation

Originally Posted by .Memo View Post
You have to follow laws of partial faction.
remember that...

Third Row correction: B/(x-1)^2

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Old 28th November 2011, 02:34 AM   #5 (permalink)
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Re: Partial Fraction of Proper Fraction's factorisation

Oh yeah... (x-1)^2 careless, no wonder i can't get the solution.


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Old 28th November 2011, 02:47 AM   #6 (permalink)
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Re: Partial Fraction of Proper Fraction's factorisation

Originally Posted by .Memo View Post
Oh yeah... (x-1)^2 careless, no wonder i can't get the solution.
No worries, you have been doing a great job helping others. Keep it up. Peace.

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Old 16th December 2011, 09:36 AM   #7 (permalink)
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Re: Partial Fraction of Proper Fraction's factorisation

thanks now I understand

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