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c4 partial fractions help

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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help
Original post by JSM1
help

Please try not to bump your thread too often. Is that supposed to be:

x3−x2−x−3x(x−1)\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:

x3−x2−x−3x(x−1)\dfrac{x^3-x^2-x-3}{x(x-1)}?


yes
Have you split it into Ax\frac{A}{x} and B(x−1)\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
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 Ax\frac{A}{x} and B(x−1)\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
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 x3−x2−x−3x(x−1)\dfrac{x^3-x^2-x-3}{x(x-1)} in the form Ax+B+Cx+Dx(x−1)Ax+B + \dfrac{Cx+D}{x(x-1)}, and then apply partial fraction to the remaining, now proper, fraction.

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