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Math Help - partial fractions (quardratic factor in denominator)

  1. #1
    Newbie helloworld101's Avatar
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    Question partial fractions (quardratic factor in denominator)

    hi am having difficulty figuring this one out can u help?

    6/(x+1)(x-1)(x-4)^2

    so far this is what i got but i'm not sure its correct....

    A/(x+1) + B/(x-1) + (Cx+D)/(x-4)^2
    considering just the frc (Cx+D)/(x-4)^2
    and letting D= -4C+E
    then

    frc(Cx+D)/(x-4)^2 = (Cx - 4C + E)/(x-4)^2


    = [C/(x-4)] + E/(x-4)^2

    therefore

    6/(x+1)(x-1)(x-4)^2 = A/(x+1) + B/(x-1) + C/(x-4) + E/(x-4)^2


    so am i going good so far? cuz when i substitute values of x am getting weird numbers.
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  2. #2
    MHF Contributor
    skeeter's Avatar
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    Quote Originally Posted by helloworld101 View Post
    hi am having difficulty figuring this one out can u help?

    6/(x+1)(x-1)(x-4)^2
    you have a repeated linear factor ...

    \frac{6}{(x+1)(x-1)(x-4)^2} = \frac{A}{x+1} + \frac{B}{x-1} + \frac{C}{x-4} + \frac{D}{(x-4)^2}

    6 = A(x-1)(x-4)^2 + B(x+1)(x-4)^2 + C(x+1)(x-1)(x-4) + D(x+1)(x-1)

    let x = -1 ...

    6 = -50A ... A = -\frac{3}{25}

    let x = 1 ...

    6 = 18B ... B = \frac{1}{3}

    let x = 4 ...

    6 = 15D ... D = \frac{2}{5}

    let x = 2 ...

    6 = 4A + 12B - 6C + 3D = -\frac{12}{25} + 4 - 6C + \frac{6}{5}

    C = -\frac{16}{75}


    \frac{6}{(x+1)(x-1)(x-4)^2} = -\frac{3}{25(x+1)} + \frac{1}{3(x-1)} - \frac{16}{75(x-4)} + \frac{2}{5(x-4)^2}
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  3. #3
    MHF Contributor red_dog's Avatar
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    \frac{6}{(x+1)(x-1)(x-4)^2}=\frac{A}{x+1}+\frac{B}{x-1}+\frac{C}{x-4}+\frac{D}{(x-4)^2}

    6=A(x-1)(x-4)^2+B(x+1)(x-4)^2+C(x+1)(x-1)(x-4)+D(x+1)(x-1)

    x=1\Rightarrow 6=18B\Rightarrow B=\frac{1}{3}

    x=-1\Rightarrow 6=-18A\Rightarrow A=-\frac{1}{3}

    x=4\Rightarrow 6=15D\Rightarrow D=\frac{2}{5}

    Now differentiate both members:

    0=A(x-4)^2+2A(x-1)(x-4)+B(x-4)^2+2B(x+1)(x-4)+

    +C(x-1)(x-4)+C(x+1)(x-4)+C(x+1)(x-1)+D(x-1)+D(x+1)

    x=4\Rightarrow 0=15C+8D\Rightarrow C=-\frac{16}{75}
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