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Math Help - another fraction to decompose

  1. #1
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    another fraction to decompose

    (x^2+2x+8) / (x^3-4x^2+4x)
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  2. #2
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    Quote Originally Posted by cityismine View Post
    Express \frac{x^2+2x+8}{x^3-4x^2+4x} as a sum of partial fractions

    first factor the denominator \frac{x^2+2x+8}{x(x-2)^2}

    \frac{x^2+2x+8}{x(x-2)^2} \equiv \frac{A}{x} + \frac{B}{x-2} + \frac{C}{(x-2)^2}


    Now continue doing this with the method your familiar with.

    Bobak
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by cityismine View Post
    (x^2+2x+8) / (x^3-4x^2+4x)
    Factor the Denominator first:

    x^3-4x^2+4x=x(x^2-4x+4)=x(x-2)^2

    Thus, the PFD is :

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

    Try taking it from here.
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  4. #4
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    2/x - 1/(x-2) + 8/(x-2)2

    Is that the right answer?
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  5. #5
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    Quote Originally Posted by cityismine View Post
    2/x - 1/(x-2) + 8/(x-2)2

    Is that the right answer?
    Yes, Good job

    Bobak
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  6. #6
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    OMG, the book was showing a different answer, so I thought it was wrong. I guess the book is wrong.
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