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Math Help - Partial Fractions

  1. #1
    Super Member
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    Partial Fractions

    Hi
    Need help on expressing the following equation in partial fractions:
    \frac{x^2-x+3}{(x^2-3x+7)(x+2)^2)}
    =\frac{A}{(x+2)}+\frac{B}{(x+2)^2}+\frac{Cx+D}{x^2-3x+7}

    A(x+2)(x^2-3x+7)+B(x^2-3x+7)+(Cx+D)(x+2)^2=x^2-x+3

    Equation 1: A+C=0
    Equation 2: -A+B+4C+D=1
    Equation 3: A-3B+4C+4D=-1
    Equation 4: 14A+7B+4D=3

    Equation 2-3
    -A+B+4C+D=1- A-3B+4C+4D=-1
    Gets: -2A+4B-3D=2

    Equation (2-3)-4
    -2A+4B-3D=2- 14A+7B+4D=3
    35B-17D=17

    Don't know what to do next??

    P.S
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  2. #2
    MHF Contributor

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    Quote Originally Posted by Paymemoney View Post
    Hi
    Need help on expressing the following equation in partial fractions:
    \frac{x^2-x+3}{(x^2-3x+7)(x+2)^2)}
    =\frac{A}{(x+2)}+\frac{B}{(x+2)^2}+\frac{Cx+D}{x^2-3x+7}

    A(x+2)(x^2-3x+7)+B(x^2-3x+7)+(Cx+D)(x+2)^2=x^2-x+3

    Equation 1: A+C=0
    Equation 2: -A+B+4C+D=1
    Equation 3: A-3B+4C+4D=-1
    Equation 4: 14A+7B+4D=3

    Equation 2-3
    -A+B+4C+D=1- A-3B+4C+4D=-1
    Gets: -2A+4B-3D=2

    Equation (2-3)-4
    -2A+4B-3D=2- 14A+7B+4D=3
    35B-17D=17

    Don't know what to do next??

    P.S
    You have eliminated A and C from your equation but have not yet used equation 1. Use that (which involves only A and C) together with the other equations to get a second equation in B and D only. Then combine the two equations to eliminate either B or D so you can solve for the remaining value.
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  3. #3
    Super Member
    Joined
    Dec 2008
    Posts
    509
    ok thanks, i'll try that
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