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Math Help - Partial Fraction Help?

  1. #1
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    Partial Fraction Help?

    I need a little help with this problem:
    8/(y^3-4y).
    The book says the answer is -2/y+1/(y-2)+1/y+2 but I'm not following how they got that answer.
    I have a feeling my main problem is that my algebra is rusty .
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  2. #2
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    Quote Originally Posted by Cbcollier91 View Post
    I need a little help with this problem:
    8/(y^3-4y).
    The book says the answer is -2/y+1/(y-2)+1/y+2 but I'm not following how they got that answer.
    I have a feeling my main problem is that my algebra is rusty .
    \frac{8}{y^3-4y}=\frac{8}{y\left(y^2-4\right)}=\frac{8}{y(y+2)(y-2)}

    Then seperate this into

    \frac{8}{y(y+2)(y-2)}=\frac{A}{y}+\frac{B}{y+2}+\frac{C}{y-2}

    Lastly, find A, B and C
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  3. #3
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    Quote Originally Posted by Cbcollier91 View Post
    I need a little help with this problem:
    8/(y^3-4y).
    The book says the answer is -2/y+1/(y-2)+1/y+2 but I'm not following how they got that answer.
    I have a feeling my main problem is that my algebra is rusty .
    \frac{8}{y(y+2)(y-2)} = \frac{A}{y} + \frac{B}{y+2} + \frac{C}{y-2}

    8 = A(y+2)(y-2) + By(y-2) + Cy(y+2)<br />

    let y = 0 ... A = -2

    let y = 2 ... C = 1

    let y = -2 ... B = 1

    \frac{8}{y(y+2)(y-2)} = -\frac{2}{y} + \frac{1}{y+2} +  \frac{1}{y-2}
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