Resolvable by radicals

Algebra Level 5

The formula $\frac{ - b \pm \sqrt{b^2 - 4ac}}{2a}$ is well known to obtain the solutions of the general equation of second degree $ax^2 + bx + c = 0$ . When it is possible to express the solutions of a polynomial equation by using only operations with its coefficients: sum, substraction, product, division and extraction of roots ( $\sqrt{ \cdot }, \space \sqrt[3]{\cdot}, ...$ ) it is said that the equation is "resolvable by radicals."

The general equation of degree n, $p(x) = a_n x^n + a_{n -1}x^{n -1} + ... + a_1 x + a_0 = 0$ is "resolvable by radicals" if and only if $n \leq a$ . What is $a$ ?

Details.-

1) $a$ is the maximum value for the general equation of degree n to be resolvable by radicals

2) $p(x) \in \mathbb{R}[x], \space a_n \neq 0$

All equations are resolvable by radicals 6 7 3 5 4

1 solution

Mark Hennings
Dec 15, 2016

Without typing out the whole of Galois Theory, the quintic $f(x) \; = \; x^5 - 10x + 5$ is irreducible over $\mathbb{Z}[x]$ with precisely three real roots, and so the Galois group of $f(x)$ over $\mathbb{Q}$ is a subgroup of $S_5$ containing an element of order $5$ (a $5$ -cycle) and a transposition, and so must be $S_5$ , which is not a soluble group. Thus the quintic equation $f(x) = 0$ cannot be solved by radicals.

Since there are (comparatively) well-known formulae for the solution of cubics and quartics by radicals (each based, ultimately, on the solubility of $S_3$ and $S_4$ respectively), the answer to the question is $\boxed{4}$ .

Thank you very much for your solution... The idea is this, the symetrical group $S_5$ is not a soluble group...It's impossible a whole solution right now , here, because there are a long way for proving all results ... For n=3 and n =4 are the Cardano-Ferrari formulas...

Guillermo Templado - 4 years, 6 months ago

Wait, so we can actually factor all polynomial that is at least the fourth degrees ?

I really need to learn more about cubic and quartic polynomial, just beyond my knowledge for this time

Jason Chrysoprase - 4 years, 6 months ago

Yes indeed. You might like to look at this page (and the link at its bottom to quartic equations), which give you methods for solving cubics and quartics.

Historically, methods for solving cubics and quartics by radicals were found quite early on (in the 1500s), and the comparative ease with which these were found led people to believe that it should be possible to solve quintics by radicals as well. What was surprising was that it took about 300 years, until the 19th century, for people to find a method of showing why the quintic was not soluble by radicals, and to understand what was truly going on.

Mark Hennings - 4 years, 6 months ago

This problem means "more or less" than there exists a general formula for solving polynomials of degree at most 4, using only the coefficients of the polynomial( over determined fields): sum, substraction, product, division and square roots see interpretation of this article ... If the polynomial has degree 5 or higher degree, there doesn't exist a general formula for solving them... This problem requires, if you want a whole proof, to have (not a few) knowledge of group theory and Galois theory...

Guillermo Templado - 4 years, 6 months ago

Furthemore, due to the fundamental theorem of algebra all polynomials of degree $n \ge 1$ , in $\mathbb{C}[x]$ can be factorized in a product of polynomials of degree 1, i.e, the only irreducible polynomials in $\mathbb{C}[x]$ are the polynomials of degree 1...

Guillermo Templado - 4 years, 6 months ago

Resolvable by radicals

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