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I am looking to find a derivation, or a intuitive explanation as for why a partial fraction with a repeated factor needs to include a factor in the expansion for each power possible. How does one derive this? $(x-1)^2$ splits into $x-1$ and $(x-1)^2$ but this contradicts what I have previously learnt, that two linear factors can be split into their seperate factors. Hence, expressing $(x-1)^2=(x-1)(x-1)$, should then splits into $x-1$ and $x-1$. Obviously this doesn't work, so why does the rule change here?

At school, they have taught us to simply remember all the general forms that one may see, and the subsequent forms of the partial fractions, without any intuition. I am looking for a way to split a rational fraction into its partial fractions without the need to memorise any of these. How can you do it?

I have looked at every single related question on stackexchange, but none address my problem.

Edit: linked answer does not answer my question. It focuses on a quadratic factor, which can be thought of as a combination of linear factors that are complex, but it does not address my question

James Chadwick
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  • https://math.stackexchange.com/questions/368665/derivation-of-the-general-forms-of-partial-fractions – RobPratt May 12 '24 at 03:52
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    This is a good question. It would take some time for me to provide a good answer, but the short version is "linear independence," in a way. – Sean Roberson May 12 '24 at 03:52
  • Consider it $$ \frac1{(x-1)^2}=\frac{A}{(x-1)}+\frac{B}{(x-1)}=\frac{A+B}{(x-1)} $$ This is mistaken since we compress a polynomial of degree $2$ into $1$ Is this your question? –  May 12 '24 at 03:58
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    "...but this contradicts what I have previously learnt..." : What exactly did you learn, that creates the contradiction, and how/where did you learn it? – user2661923 May 12 '24 at 04:09
  • Case (3) in the linked question deals with repeated factors, which seems to be exactly what you're asking about. Anyway, this has been asked so many times before on this site that surely there must be some answer that works... Did you look at all these: https://math.stackexchange.com/q/20963/1242, https://math.stackexchange.com/q/185264/1242, https://math.stackexchange.com/q/48051/1242, https://math.stackexchange.com/q/3132606/1242, https://math.stackexchange.com/q/2108566/1242, etc.? – Hans Lundmark May 12 '24 at 05:13

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