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Given a polynomial equation (with real coefficients of any degree with any number of repeating roots), let say

$x^5 + 6x^4 - 18x^3 - 10x^2 + 45x - 24 = 0$, ... (A)

it can be written as

$(x-1)^2 (x+8) (x^2-3) = 0$.

As you can see, it has a repeating root $x = 1$.

Forming a new polynomial by removing the duplicate root (i.e, taking repeating roots only once), we will get

$(x-1) (x+8) (x^2-3) = 0$.

Then expanding it, we get

$x^4+7 x^3-11 x^2-21 x+24 = 0$. ... (B)

Is it possible to get a polynomial having all of the roots of the given polynomial but taking the repeating roots only once, i.e, getting (B) from (A), directly without finding any roots? If possible, how to do the same?

Bill Dubuque
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Bhaskar
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  • How does one know that there is a repeated root without knowledge of any root? – Mark Viola Apr 10 '15 at 17:11
  • It is a polynomial condition on the coefficients of the polynomial, also known as $Res(f,f')=0$. @Dr.MV cfr Resultant –  Apr 10 '15 at 17:16
  • @Dr.MV : I don't know. This question came to my mind but I could not think of a way to do it. I am expecting that the algorithm/solution should return the same polynomial if it has no repeating roots. – Bhaskar Apr 10 '15 at 17:20
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    The squarefree part is $, f/\gcd(f,f'),$ over a field of characteristic 0 – Bill Dubuque Apr 10 '15 at 17:25
  • @G.Sassatelli Does that tell one whether the original polynomial of interest has multiple roots? And how does one proceed to find the "resultant" without any knowledge of any root? – Mark Viola Apr 10 '15 at 17:26
  • @BillDubuque's idea is the way to go, I was about to write it as an answer when he posted the comment. Differentiating, taking $\gcd$, and dividing polynomials are all operations that you can do much quicker than factoring a polynomial. – Nate Apr 10 '15 at 17:28
  • So gcd can also remove the roots if they repeat more than 2 times, i.e., if the polynomial has a factor having powers of more than 2? @BillDubuque – Bhaskar Apr 10 '15 at 17:32
  • What if the repeating roots are imaginary? @BillDubuque – Bhaskar Apr 10 '15 at 17:33
  • @Dr.MV It does; for instance, the proof should be somewhere in S. Lang's "Algebra".

    The resultant of two polynomials $p,\ q$ is the determinant of the Sylvester matrix $S_{p,q}$.

    $Res(f,f'):=\det S_{f,f'}$

    As you can see, though, the inherency of our exchange is fading.

     –  Apr 10 '15 at 17:38
  • @L16H7 Because $,f,$ is real, the nonreal roots must occur in conjugate pairs, so the repeated nonreal roots combine to yield a repeated real quadratic factor (the minimal poly of the nonreal root). – Bill Dubuque Apr 10 '15 at 17:43
  • @BillDubuque : You may write an answer. Then I can mark it. – Bhaskar Apr 10 '15 at 17:49

1 Answers1

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The square free kernel (or radical) of a polynomial, i.e. the product of all prime = irreducible factors (up to a constant factor) can be computed efficiently via a gcd calculation, namely

$$ {\rm rad}(f(x))\, =\, \frac{f(x)}{\gcd(f(x),f'(x))}\qquad$$

This works because taking the dervative decrements the multiplicity of each prime factor, thus taking the above quotient has the effect of cancelling out repeated factors from $\,f.\,$ (Beware that this need not work over fields of characteristic $\neq 0).$

Iterating we can efficiently compute a complete squarefree decomposition using only gcds and derivatives (no need for more expensive general factorization algorithms), e.g. see Yun's algorithm.

A typical example is Hermite's method for integrating rational functions.

Contrast this with the number (integer) case. Lacking an analog of the derivative for numbers we cannot use the same idea. It is trivial to to compute the radical and square/squarefree parts if we are given the prime factorization. Currently we do not know any better way, and it is widely suspected that there is none, i.e computing the squarefree part of an integer may prove to be equivalent to factorization.

This problem is important because one of the main tasks of computational algebraic number theory reduces to it (in deterministic polynomial time). Namely the problem of computing the ring of integers of an algebraic number field depends upon the square-free decomposition of the polynomial discriminant (when computing an integral basis). See this answer for further elaboration.

Bill Dubuque
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