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    express the following improper fraction as a partial fraction:

    x^3-x^2-x-3/x (x-1)
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    help
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    (Original post by JSM1)
    help
    Please try not to bump your thread too often. Is that supposed to be:

    \dfrac{x^3-x^2-x-3}{x(x-1)}?
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    yes
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    (Original post by Farhan.Hanif93)
    Please try not to bump your thread too often. Is that supposed to be:

    \dfrac{x^3-x^2-x-3}{x(x-1)}?
    yes
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    Have you split it into \frac{A}{x} and \frac{B}{(x-1)}? And then equated that to the original.
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    no. is that not an improper fraction so you are meant to divide the denominator by the numerator
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    (Original post by JSM1)
    yes
    if you're unsure where to start, try performing long division first to express it in terms of a quotient and a remainder, and then use partial fractions on the remainder. If you're unsure how to do this, throw up some working.

    (Original post by uponthyhorse)
    Have you split it into \frac{A}{x} and \frac{B}{(x-1)}? And then equated that to the original.
    This does not work immediately for improper fractions.
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    it says it's an improper fraction however you can still split it into a/x + b/x-1. This gives you the second half of the answer but the full answer is

    x+3/x-4/x-1
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    actually don't worry I got the full answer. there was no factor so you could just split it into partial fractions
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    (Original post by JSM1)
    it says it's an improper fraction however you can still split it into a/x + b/x-1.
    Not true. This only works if the order of the numerator is lower than that of the denominator. In this case, one would require the numerator to be linear or constant (as the denominator is quadratic) but it is instead cubic.

    First use long division to express \dfrac{x^3-x^2-x-3}{x(x-1)} in the form Ax+B + \dfrac{Cx+D}{x(x-1)}, and then apply partial fraction to the remaining, now proper, fraction.
 
 
 
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Updated: September 22, 2017
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