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Thread: Partial Fractions

  1. #1
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    Partial Fractions

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

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

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

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

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

    Don't know what to do next??

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

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

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

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

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

    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
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    ok thanks, i'll try that
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