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I know that an exercise like this \begin{gather*} \frac{5x^2 - 12x + 6}{(x - 2) (x - 1) (2 x - 3)} \end{gather*} is resolvable with several approaches the Heaviside "Cover-up method" e.g. \begin{align*} \frac{5x^2 - 12x + 6}{(x - 2) (x - 1) (2 x - 3)} &= \frac{5x^2 - 12x + 6}{2(x - 2) (x - 1) (x - 3/2)} \\[1ex] &= \frac{A}{x - 2} + \frac{B}{x - 1} + \frac{C}{x - 3/2} \end{align*} to estimate the unknown constant "A" appling the rule: \begin{gather*} A = \left. \frac{5x^2 - 12x + 6}{2 (x - 1) (x - 3/2)} \right|_{x = 2} = \frac{5 (2)^2 - 12 (2) + 6}{2 ((2) - 1) ((2) - 3/2)} = 2 \end{gather*} It works fine, but I'm interested to understand if is possible use the modulo with an inverted function like this: \begin{align*} A &= \left. \frac{5x^2 - 12x + 6}{2 (x - 1) (x - 3/2)} \right|_{x - 2 = 0} \end{align*} In the follow I'm looking for the unknown constant $\alpha$: \begin{gather*} (x - 2)(x - \alpha) = x^2 - (\alpha + 2) x + 2\alpha = 0\qquad \Longrightarrow \qquad x^2 - (\alpha + 2) x = - 2\alpha \\[1ex] x^2 - (\alpha + 2) x \equiv x^2 - \frac{12}{5}x \qquad \Longrightarrow \qquad - (\alpha + 2) = - \frac{12}{5} \qquad \Longrightarrow \qquad \alpha = \frac{2}{5} \\[1ex] -10\alpha = 5 \left( x^2 - (\alpha + 2) x \right) = 5x^2 - 12x \qquad \Longrightarrow \qquad 5x^2 - 12x = -4 \end{gather*} and so: \begin{align*} A &= \frac{-4 + 6}{2 (x - 1) (x - 3/2)} = \frac{1}{(x - 1) (x - 3/2)} \pmod{(x - 2)} \end{align*} At this point I don't know to proceed or I have mistraked.

Could you improve the resolution? Thanks a lot.

user-632306
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  • It's simpler to first clear denominators before modding out (i.e. evaluating via the polynomial remainder theorem), e.g. see here, which also shows how to extend Heaviside to nonlinear denominator factors. – Bill Dubuque Apr 17 '23 at 21:49
  • Calculating polynomials mod x-2 or evaluating them at 2 is essentialy the same (just a bit longer). You did it for the numerator, you can do the same for the denominator. – Anne Bauval Apr 17 '23 at 21:50
  • It's not clear exactly why you are stuck. Are you using a method that evaluates via the polynomial remainder theorem (i.e. $f(a) = f(x)\bmod x!-!a),,$ and you are not sure how to apply that to quotients of polynomials ("rational functions")? [this is what I guessed in my first comment]. – Bill Dubuque Apr 17 '23 at 22:03
  • @Bill Dubuque - Unfortunately I'm stuck on polynomial remainder theorem. Thank you very much for the link I'm studying your answer to the other question. – user-632306 Apr 18 '23 at 09:39
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    @Anne Bauval - Thanks for your suggestion, I don't know how resolve with the modulo (x - 2) and I have try to solve it in that way to understand the difficulty. – user-632306 Apr 18 '23 at 09:52

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