How to tell when a polynomial is solvable by radicals

evene · 2016-10-14 16:04:38

I was just wondering about this, but is there a way to tell if a polynomial of degree n is solvable radicals? For example, is the polynomial

and six degree that can be solved by radicals? And if so, what would be its solutions?

Yea, this is kind of a two part question. Hope you guys are fine with that?

bobbym · 2016-10-14 19:41:35

The roots of a big polynomial are a rather difficult thing. Most people do not appreciate when a CAS spits out

{-1.50446, -1.4034, -0.592903, 1.57365, -0.0364475 - 1.4827 I, -0.0364475 + 1.4827 I}

just how hard that was to do. Generally, Abel proved a long time ago there was no general method to get the roots of an equation in radicals for polynomials of degree higher than 4. This confuses people into thinking we can never get the roots. The theorem as far as I know only states that there is no general formula like the quadratic equation for polynomials of degree higher than 4.

Of course, polynomials are our favorite form for approximation and we know more about them than anything else.

zetafunc · 2016-10-14 21:30:45

As bobbym said, for a quintic (or higher degree) polynomial, there does not exist a formula for the roots in terms of radicals of the polynomial's co-efficients in general. More precisely, such a polynomial is solvable by radicals (over the rationals, or more generally any field with characteristic zero) if and only if its Galois group is a solvable group.

Last edited by zetafunc (2016-10-15 08:36:02)

bobbym · 2016-10-14 21:35:01

Hi zetafunc;

If you have time can you show an example?

evene · 2016-10-15 07:19:04

@Zetafunc

Does the polynomial I asked have a solvable galois group? And also, what's a galois group. I haven't gotten that far in math yet.

zetafunc · 2016-10-15 09:18:10

Computing Galois groups of algebraic fields is an extremely difficult problem in general, especially for your example. Magma says your equation has Galois group of order 72 (and it gives some permutation relations), and there are 50 groups of order 72, up to isomorphism. I don't know which group it is for your example. However, all groups of order 72 are solvable (see Burnside's theorem with 72 = 2^3 * 3^2, i.e. (p,q,a,b) = (2,3,3,2)), so in particular your Galois group is solvable and so your polynomial is solvable by radicals -- that is to say that the roots bobbym found in post #2 can be given explicitly by radicals, but actually finding some closed form expression for those roots is very, very hard. (But they do exist.)

Firstly, do you know any group theory? Without going into details about field extensions and automorphisms, the simplest explanation I can give is that the Galois group of a polynomial is basically something which permutes its roots around. For instance, it is well-known that if you know a root of a polynomial, then its complex conjugate is also a root. This is no coincidence: the reason for this is that complex conjugation is a field automorphism (which are basically the things which live inside Galois groups).

I'll do some simple examples like x^n - 1 = 0 later.

Last edited by zetafunc (2016-10-15 09:44:19)

evene · 2016-10-16 03:15:20

Well, I do know very little about group theory. I understand about fields and groups and maybe some permutations?

Hm.. then what are the roots of the six degree? I need it to solve an equation!

evene · 2016-10-16 03:16:44

Actually, I wonder if it's possible to factor it into two cubics, and solving for the roots of the cubics generate the roots of the six degree... hm...

zetafunc · 2016-10-16 03:33:48

Mathematica says your roots are approximately:

bobbym wrote:

{-1.50446, -1.4034, -0.592903, 1.57365, -0.0364475 - 1.4827 I, -0.0364475 + 1.4827 I}

and those approximations produce

Multiplying through by 3 gives you something close to your original polynomial.

Last edited by zetafunc (2016-10-16 04:03:46)

zetafunc · 2016-10-16 03:58:08

evene wrote:

Actually, I wonder if it's possible to factor it into two cubics, and solving for the roots of the cubics generate the roots of the six degree... hm...

It's possible to factorise it into monic irreducibles if you wanted to, over . But that isn't easy.

evene · 2016-10-16 04:35:57

Yeah, is it possible to get the exact values, if possible? The roots of this are crucial to finding an unknown in something I'm working on.

bobbym · 2016-10-16 04:38:19

The only software I have in my possesion can only do that for a quintic. Yours is a sextic. That means one root has to be found exactly and deflated out. This could be impossible to do.

evene · 2016-10-16 04:44:56

Dang it...

zetafunc · 2016-10-16 04:48:58

evene wrote:

Yeah, is it possible to get the exact values, if possible? The roots of this are crucial to finding an unknown in something I'm working on.

They do exist, but I don't know of any way to get the nice closed form that you want. That is an extremely difficult problem in general -- even determining whether a polynomial is solvable by radicals is difficult in general, let alone finding its roots.

Magma says the Galois group of your polynomial has size 72. All groups of this size are soluble, and therefore your polynomial is solvable by radicals. But I can't give you any more information on how to find those -- in mathematics, very often we can prove the existence of something without being able to find it.

Last edited by zetafunc (2016-10-16 04:50:17)

evene · 2016-10-16 05:28:22

@Zetafunc

Actually, the sextic can be factored into

Now, I just need to find the roots of the cubics...

bobbym · 2016-10-16 05:36:03

Hi;

I can get the two imaginary roots exactly and then I can maybe deflate them out to form a quartic. Quartics have a general formula.

zetafunc · 2016-10-16 05:36:13

You can use the cubic formula for that.

evene · 2016-10-16 05:46:20

What is the formula for all three roots of a cubic? After searching the internet for some time, I only get one formula for the general cubic...

And also, where has Bob Bundy been? Haven't seen him in a long time..

Last edited by evene (2016-10-16 05:46:48)

bobbym · 2016-10-16 05:48:33

It is posted right here in this forum.

Bob was in here earlier.

zetafunc · 2016-10-16 05:51:36

See (54), (55) and (56) here.

Alternatively you can divide your cubic by a monic irreducible once you've found a root, then you'll get a quadratic you can solve with the quadratic formula.

evene · 2016-10-16 06:04:49

I am getting

as one of the roots, is that right?? I don't trust my math.

Last edited by evene (2016-10-16 06:05:09)

bobbym · 2016-10-16 06:08:28

If you can not get the closed forms I have all 6 roots in closed form if you need them.

Math Is Fun Forum

#1 2016-10-14 16:04:38

How to tell when a polynomial is solvable by radicals

#2 2016-10-14 19:41:35

Re: How to tell when a polynomial is solvable by radicals

#3 2016-10-14 21:30:45

Re: How to tell when a polynomial is solvable by radicals

#4 2016-10-14 21:35:01

Re: How to tell when a polynomial is solvable by radicals

#5 2016-10-15 07:19:04

Re: How to tell when a polynomial is solvable by radicals

#6 2016-10-15 09:18:10

Re: How to tell when a polynomial is solvable by radicals

#7 2016-10-16 03:15:20

Re: How to tell when a polynomial is solvable by radicals

#8 2016-10-16 03:16:44

Re: How to tell when a polynomial is solvable by radicals

#9 2016-10-16 03:33:48

Re: How to tell when a polynomial is solvable by radicals

#10 2016-10-16 03:58:08

Re: How to tell when a polynomial is solvable by radicals

#11 2016-10-16 04:35:57

Re: How to tell when a polynomial is solvable by radicals

#12 2016-10-16 04:38:19

Re: How to tell when a polynomial is solvable by radicals

#13 2016-10-16 04:44:56

Re: How to tell when a polynomial is solvable by radicals

#14 2016-10-16 04:48:58

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#15 2016-10-16 05:28:22

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#16 2016-10-16 05:36:03

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#17 2016-10-16 05:36:13

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#18 2016-10-16 05:46:20

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#19 2016-10-16 05:48:33

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#20 2016-10-16 05:51:36

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#21 2016-10-16 06:04:49

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#22 2016-10-16 06:08:28

Re: How to tell when a polynomial is solvable by radicals

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