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Math Help - Decompose into partial fractions

  1. #1
    Junior Member Greymalkin's Avatar
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    Decompose into partial fractions

    {3x^2-11x-26} \over {(x^2-4)(x+1)}

    Book answer is:
    {-3\over (x-2)}+{2\over (x+2)}+{4\over (x+1)}

    My answer is:
    {-1\over (x-2)}+{-10\over (x+2)}+{4\over (x+1)}

    I seem to get the same answer for C, but my system of equations is off:
    (Ax+B)(x+1)+C(x^2-4)
    Ax^2+Ax+Bx+B+Cx^2-4C
    (A+C)x^2+(A+B)x+(B-4C)

    3=A+C
    -11=A+B
    -26=B-4C

    Reverse engineering the solutions shows my system to be incorrect, which seems wierd because I followed a text example step by step.
    Why am I wrong?
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  2. #2
    MHF Contributor

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    Re: Decompose into partial fractions

    Quote Originally Posted by Greymalkin View Post
    {3x^2-11x-26} \over {(x^2-4)(x+1)}

    Book answer is:
    {-3\over (x-2)}+{2\over (x+2)}+{4\over (x+1)}
    My answer is:
    {-1\over (x-2)}+{-10\over (x+2)}+{4\over (x+1)}
    seem to get the same answer for C, but my system of equations is off:
    (Ax+B)(x+1)+C(x^2-4)
    Ax^2+Ax+Bx+B+Cx^2-4C
    (A+C)x^2+(A+B)x+(B-4C)
    You work with A(x+2)(x+1)+B(x-2)(x+1)+C(x^2-4).
    Thanks from Greymalkin
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  3. #3
    Junior Member Greymalkin's Avatar
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    Re: Decompose into partial fractions

    Wasn't really getting the gist of decomposition until I watched a khan academy video on it, which sort of explained the relatedness of the method. But it was your post that made the lightbulb go on, thank you very much!
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