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I have searched all over to get an idea of which method is preferrable for computers to solve the real roots of a low degree polynomial, specifically a quintic, but have not come up with a simple answer.

Is there a general algorithm which is considered to be the fastest/most efficient for such a low degree?

Gary Allen
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    Yes, there are several numerical methods. This site has also several posts on it (I didn't look yet for more), for example here. In some cases, one can also apply algebraic methods, see for example this post. Also interesting is the wikipedia article. – Dietrich Burde Jan 28 '21 at 12:38
  • How does this https://arxiv.org/pdf/1209.3555.pdf method sound? I've just come across it. @DietrichBurde. Also, if I'm only concerend with a certain interval (say [0, 1]), is there a more efficient way? Also, a lot of these methods seem "overkill" as they cater for such a high degree. Is there a method that can be used specifically for low degrees and is faster? – Gary Allen Jan 28 '21 at 12:41
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    What's wrong with simple numerical techniques like Newton's method? – Allawonder Jan 28 '21 at 13:09
  • I plan on using a combination of Newton's method and the bisection method, however they required either a good initial guess or for you to isolate the roots. I'm still not too sure which method to use for root isolation, which is sort of the foundation of my question (I asked about it here too: https://math.stackexchange.com/questions/4001747/most-efficient-method-for-quintic-real-root-isolation/4001949#4001949). I'm currently looking into this https://arxiv.org/pdf/1209.3555.pdf algorithm. @Allawonder – Gary Allen Jan 28 '21 at 13:44
  • @GaryAllen By isolating roots do you mean looking separating the real line into disjoint intervals such that there exists not more than one root in each interval? – Allawonder Jan 28 '21 at 21:27
  • @Allawonder yes – Gary Allen Jan 29 '21 at 06:48
  • @GaryAllen Do you have a specific quintic in mind, or you would like to do this more generally? – Allawonder Jan 29 '21 at 06:59
  • @Allawonder my algorithm must be able to cater for all quintics, so it should be general – Gary Allen Jan 29 '21 at 08:08

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The easiest approach, if it works, is to use the Rational Root Theorem to find factors of the original quintic, reducing it to a quartic, a cubic, or quadratic.

User Yuri Negometyanov of MSE offered me a unique solution here to a problem I had with a quintic equation that resulted in multiplying $\space ABC \space $ of a Pythagorean triple. The method defined clear search limits to be used in finding a quadratic divisor $\space C\space $ of the the quintic equation, leaving a cubic that could be solved by other means. The original quintic equation was

$$ABC=2mk^5- 2 m^5k + P=0 \quad\text{ where}\quad C=m^2+n^2$$

The objective was to find $\space C:\space C|P\space $ and then to find integer solutions to $\space n=\sqrt{C-m^2},\space m\lt\sqrt{C}.\quad$ From there, we have a quadratic "factor" of the quintic. Your particular quadratic may be different.

To be brief, the limits were \begin{equation} \sqrt[3]{2P} < C < \frac{\sqrt{(4P)^{\frac{4}{5}}+1 } + 1}{2}\quad \land \quad C\bigg |\frac{P}{12} \end{equation}

and the search for this involved for $P=192720$

$$P=192720\implies \lfloor\sqrt[3]{2(192720)}\rfloor = 72 \le C \le \bigg\lfloor\frac{\sqrt{\big(4(192720)\big)^{\frac{4}{5}}+1 } + 1}{2}\bigg\rfloor=113$$ $$\land\quad \frac{P}{12} =\frac{192720}{12}=16060$$ Only two of $16060$ are between $72$ and $113$ and only one of them works properly with to yield a "solvable" cubic.

poetasis
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