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Math Help - Homework Help/Verify solution

  1. #1
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    Homework Help/Verify solution

    Would very much appreciate verification of solution to partial fraction decomposition question.

    Find the partial fraction decomposition of x^4/(x-1)^3

    Since degree of numerator is greater than denominator ... long division gives
    3x^3-3x^2+x. Since degrees are same, another long division gives:
    6x^2 -8x +3.

    With A/(x-1) + B/(x-1)^2 + C/(x-1)^3

    My solution is: A=6, B=4 and C=1

    But my "check" does not compute.

    Appreciate any help or redirection.

    Doris
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    I'm not a close friend of partial fractions but I'll do this as follows:

    \frac{{x^4 }}<br />
{{(x - 1)^3 }} = x \cdot \left( {\frac{x}<br />
{{x - 1}}} \right)^3 .

    Now

    \frac{x}<br />
{{x - 1}} = \frac{{x - 1 + 1}}<br />
{{x - 1}} = 1 + \frac{1}<br />
{{x - 1}}.

    And

    \left( {\frac{x}<br />
{{x - 1}}} \right)^3 = 1 + \frac{3}<br />
{{x - 1}} + \frac{3}<br />
{{(x - 1)^2 }} + \frac{1}<br />
{{(x - 1)^3 }}.

    Multiply this last result by x and you'll get the desired partial fraction.

    (Of course someone else will be give you a solution which involves your procedure.)
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  3. #3
    Senior Member
    Joined
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    Melbourne
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    I'm not a close friend of partial fractions but I'll do this as follows:



    Now



    And



    Multiply this last result by and you'll get the desired partial fraction.

    (Of course someone else will be give you a solution which involves your procedure.)
    THAT WAS PURE AWESOME

    Hmmm... now I need to say something relevent to stop this post being spam.
    But my "check" does not compute.
    I think that the problem is in your checking because my check worked fine:

    A(x-1)^2+B(x-1)+C = 6x^2-8x+3
    6(x^2-2x+1)+4(x-1)+1 = 6x^2-8x+3
    6x^2-12x+6+4x-4+1 = 6x^2-8x+3
    6x^2-8x+3 = 6x^2-8x+3

    Your method all looks sound.
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