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Math Help - partial fraction

  1. #1
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    Sep 2007
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    partial fraction

    need help again in express the attached equation in partial fraction. the power 3 in the denominator seems to bug me again.



    thanks!
    Last edited by samtrix; October 3rd 2007 at 11:41 PM. Reason: oops!
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  2. #2
    MHF Contributor
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    (7 +10x -x^2) / (x^3 +3x^2 -x -3)

    The denominator,
    x^3 +3x^2 -x -3
    = (x^3 +3x^2) -(x +3)
    = (x^2)(x +3) -(x +3)
    = (x +3)(x^2 -1)
    = (x +3)(x +1)(x -1)

    So,
    (7 +10x -x^2) / (x^3 +3x^2 -x -3) = A/(x+3) +B/(x+1) +C/(x-1)
    Multiply both sides by (x+3)(x+1)(x-1),
    7 +10x -x^2 = A(x+1)(x-1) +B(x+3)(x-1) +C(x+3)(x+1) -----------(i)

    When x = -1,
    7 +10(-1) -(-1)^2 = 0 +B(-1+3)(-1-1) +0
    -4 = B(2)(-2)
    B = 1 ---------------**

    When x = 1,
    7 +10(1) -(1)^2 = 0 +0 +C(1+3)(1+1)
    16 = C(4)(2)
    C = 2 ---------------**

    When x = -3,
    7 +10(-3) -(-3)^2 = A(-3+1)(-3-1) +0 +0
    7 -30 -9 = A(-2)(-4)
    -32 = 8A
    A = -32/8 = -4 -------**

    Therefore,
    (7 +10x -x^2) / (x^3 +3x^2 -x -3) = -4/(x+3) +1/(x+1) +2/(x-1)
    Or,
    (7 +10x -x^2) / (x^3 +3x^2 -x -3) = 1/(x+1) +2/(x-1) -4/(x+3) -------------answer.
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  3. #3
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    Sep 2007
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    thanks ticbol! been a great help!
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