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Math Help - Partial Fraction Decomposition

  1. #1
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    Partial Fraction Decomposition

    Partial Fraction Decomposition of the Following:

    (4x^2-8x+48)/(x^4-8x^3+16x^2)<br />
    I factored out and simplified the denominator which gives me: x^2(x-4)(x-4).

    From that point on I write out the decomposition relative to A,B,C,D and the factored out denominator which is: (A/x^2)+(B/x)+(C/x-4)+(D/x-4).

    I need to find the constants A B C & D that equal the original numerator.
    In other words:
     (A/x^2)+(B/x)+(C/x-4)+(D/x-4)= (4x^2-8x+48)/(x^4-8x^3+16x^2)

    from the decomposition relative to A B C and D, I combine all terms under one single denominator and get:

    (Ax^2-8Ax+16A+Bx^3-8Bx^2+16Bx+Cx^3-4Cx^2+Dx^3-4Dx^2)/x^2(x^2-8x+16)

    I create systems of equations according to the degree of each variable and get:

    B+D+C=0

    A-8B-4C-4D=4

    -8A+16B=-8

    16A=48

    When I solve the system I get no solutions?
    Can anyone tell me where I am going wrong?
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  2. #2
    Senior Member BAdhi's Avatar
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    Quote Originally Posted by dagbayani481 View Post
    From that point on I write out the decomposition relative to A,B,C,D and the factored out denominator which is: (A/x^2)+(B/x)+(C/x-4)+(D/x-4).
    the decomposition should be \frac{4x^2-8x+48}{x^4-8x+16x^2}=\frac{A}{x^2}+\frac{B}{x}+\frac{C}{(x-4)}+\frac{D}{(x-4)^2}
    Last edited by BAdhi; February 3rd 2011 at 01:01 AM. Reason: latex error resolved
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  3. #3
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    This is odd but from my end I'm seeing a LaTeX Error. Can't see what you wrote ):

    Edit: Ah, I see it now. Thank you so much!
    Last edited by dagbayani481; February 3rd 2011 at 01:07 AM.
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