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Message boards : General discussion : Root(s) of a cubic polynomial

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HellGauss
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Message 77549 - Posted: 24 Jun 2014 | 21:19:51 UTC

Just a simple question (if you know the answer...)

Is it always possible to decompose a cubic real polynomial in 3 factors of degree one each or in two factor of degree one and two, using only basic operations and n-roots?

I know about Cardano's formula, but my question is a little bit more subtle.

If the polynomial has only one real root, you can find it using Cardano's formula. Then you can easily find the other polynomial by division.

But if you have three real roots, in Cardano's formula you need to calculate the cubic root of a complex number, which, as far as I know, need trigonometric functions to obtain explicit values of the real and imaginary part.

Thanks for any insight on this problem
HG
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Message 77577 - Posted: 25 Jun 2014 | 17:36:47 UTC - in response to Message 77549.

Just a simple question (if you know the answer...)

Is it always possible to decompose a cubic real polynomial in 3 factors of degree one each or in two factor of degree one and two, using only basic operations and n-roots?

Just read the Wiki article. The second sentence in [1] gives you the definitive answer. The test for existence of rational roots is straightforward, and then if you don't have them, then you have casus irreducibilis and you cannot express the roots in terms of real radicals.

HellGauss
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Message 77600 - Posted: 25 Jun 2014 | 22:10:43 UTC - in response to Message 77577.

Thanks a lot for the reference!
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Message boards : General discussion : Root(s) of a cubic polynomial

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