Admittedly, this may not be a research level question, but I am deeply curious about this.

Let $f(x) \in \mathbb{R}[x]$, and write $d = \deg f$. It is well known that if $\deg f > 2$, then $f$ is reducible over $\mathbb{R}$. This fact can easily be proved with the fundamental theorem of algebra. Indeed, by the fundamental theorem of algebra, $f(x)$ splits over $\mathbb{C}$, and since $f(x) = \overline{f(x)}$ when $x$ is real, it follows that the linear factors of $f$ must be real or come in conjugate pairs. Therefore, irreducible polynomials over $\mathbb{R}$ are only of degree 1 or 2, and it is easy to find examples of degree two polynomials with real coefficients which are irreducible.

This fact has a profound impact on the theory of partial fractions, a staple in first year calculus. Indeed, Calculus II (at least at my university) tends to spend an inordinate amount of time on 'integration' via the method of anti-derivatives, which I believe most mathematicians know is ineffective in solving the majority of problems, considering most functions do not have elementary anti-derivatives. However, in the context of rational functions, anti-differentiation and partial fractions completely solves the problem, as any rational function can be written as the sum of simpler rational functions, each with an elementary anti-derivative.

However, this fact is (I believe) far from obvious if you do not know that the only irreducible polynomials over $\mathbb{R}$ are linear or quadratic. That said, it seems to me that the method of partial fractions is much older than the fundamental theorem of algebra.

So when did people know (perhaps before the first proof of the fundamental theorem of algebra) that the only irreducible polynomials over $\mathbb{R}$ are linear or quadratic? When was the 'complete' solution of anti-derivatives of rational functions obtained? Is this history accounted for anywhere?

I apologize in advance if this question is trivial.

  • 8
    $\begingroup$ If someone knew long ago that all polynomials over the reals can be factored into polynomials of degree $\leq2$, then all that would be needed to get the fundamental theorem of algebra would be that real quadratics have complex roots. That would be known pretty much immediately after complex numbers are invented. $\endgroup$
    – Andreas Blass
    Aug 2, 2015 at 2:58
  • 3
    $\begingroup$ As I understand it, one of the reasons proving the FTA was important was to ensure that partial fractions would always work, at least in principle; I do not believe that it was a known fact before then. $\endgroup$
    – Arturo Magidin
    Aug 2, 2015 at 4:39
  • 10
    $\begingroup$ This question would be more suitable on the History of Science and Mathematics stackexchange site. You are essentially asking who first proved the fundamental theorem of algebra. That is generally attributed to Gauss, in his doctoral thesis, although I believe his argument had some topological gaps. $\endgroup$
    – KCd
    Aug 2, 2015 at 6:34
  • 3
    $\begingroup$ I wrote an essay about the mathematics of integrating rational functions in this 15 April 2006 sci.math post, and in my brief historical comments I made a reference to the MacTutor History of Mathematics archive for The fundamental theorem of algebra, which seems to be more focused on what you're looking for than the Wikipedia article. See also the comments at the end of my sci.math post about Leibniz. $\endgroup$ Aug 3, 2015 at 19:16
  • 2
    $\begingroup$ @Alex Kruckman: It took several searches in the google sci.math archive, but I finally found it. Here's google's sci.math archive of the 15 April 2006 sci.math post. BTW, the reason I included the disclaimer "then so too will be the coefficients involved in (at least one of) the antiderivative(s)" in a comment near the beginning is that any constant can be added to an antiderivative to produce another antiderivative, and thus one could add something like $\pi$ to an antiderivative, which would spoil what I was saying. $\endgroup$ Dec 17, 2022 at 15:51


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.