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Math Help - Partial Fractions problem

  1. #1
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    Partial Fractions problem

    Good Day,

    I came across this partial fractions problem yesterday and I'm confused as to why I get the wrong answer although the factorising seems correct to me. I'll list down what I did and what is the answer provided by the book.

    The steps I took to break it up
    x / (1-x^2)(x-1)
    = x / (1+x)(1-x)(x-1)
    = -x / (x-1)^2 (1+x)
    = - [1/4(x-1)] - [1/2(x-1)^2] + [1/4(1+x)]

    However, the book states that the answer is [1/4(1-x)] - [1/2(1-x)^2] + [1/4(1+x)].

    When I tested my answer and the book's, I found out that the book's answer gave the fraction that I was asked to break up.

    Is there a reason why I must factorise the denominator into (1-x)^2 (1+x) and not (x-1)^2 (1+x)?

    Thanks in advance.
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  2. #2
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    Quote Originally Posted by dd86 View Post
    Good Day,

    I came across this partial fractions problem yesterday and I'm confused as to why I get the wrong answer although the factorising seems correct to me. I'll list down what I did and what is the answer provided by the book.

    The steps I took to break it up
    x / (1-x^2)(x-1)
    = x / (1+x)(1-x)(x-1)
    = -x / (x-1)^2 (1+x)
    = - [1/4(x-1)] - [1/2(x-1)^2] + [1/4(1+x)]

    However, the book states that the answer is [1/4(1-x)] - [1/2(1-x)^2] + [1/4(1+x)].

    When I tested my answer and the book's, I found out that the book's answer gave the fraction that I was asked to break up.

    Is there a reason why I must factorise the denominator into (1-x)^2 (1+x) and not (x-1)^2 (1+x)?

    Thanks in advance.
    hi

    you are correct and the book is also correct. You just play with the algebra here.

    -1/4(x-1)= -1/4(-(1-x))=1/4(1-x)

    It doesn't matter since both forms are correct.
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  3. #3
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    (1- x)^2= (-(x-1))^2= (-1)^2(x-1)^2= (x-1)^2
    Last edited by mr fantastic; May 31st 2010 at 02:23 PM. Reason: Added [/math] tag.
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  4. #4
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    Great! Thanks for the quick response, everyone!
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