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When decomposing into a partial fraction, why does the highest degree of the numerator have to be one lower than the numerator?

For example:

$\frac{x}{x^3-1} = \frac{a}{x-1} + \frac{bx+c}{x^2+x+1}$

and not:

$\frac{x}{x^3-1} = \frac{a}{x-1} + \frac{b}{x^2+x+1}$

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  • $\begingroup$ Who claimed that? It doesn't have to be like that. $\endgroup$
    – José Carlos Santos
    Commented Nov 5, 2022 at 11:03
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    $\begingroup$ In your example, there would be no way to get rid of $x^2$ term otherwise $\endgroup$
    – Vasili
    Commented Nov 5, 2022 at 11:04
  • $\begingroup$ In the case $\frac{2x^2+x}{x^3-1}$ for example , the second approach would give $a=2$ , $a+b=1$ , $a-b=0$ , which has no solution. $\endgroup$
    – Peter
    Commented Nov 5, 2022 at 11:07
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    $\begingroup$ From the theoretical standpoint, all we know about decomposition is that degree of numerator is less than degree of the denominator so we have to assume that it can be a polynomial of degree 1 if denominator is quadratic. Similarly, if denominator is cubic (root multiplicity of three), we have to assume that numerator can be quadratic. $\endgroup$
    – Vasili
    Commented Nov 5, 2022 at 11:20
  • $\begingroup$ Try it, in your second expression you will find that $b$ depends on $x$ i.e you will find different values of $b$ depending on which $x$ you use to find it. $\endgroup$
    – Paul
    Commented Nov 5, 2022 at 12:29

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