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Showing a polynomial is not solvable by radicals

  1. Nov 30, 2016 #1
    1. The problem statement, all variables and given/known data
    Show that the polynomial f(x)=x^5-3x^4+6x^3+18x^2-3 is NOT solvable by radicals

    2. Relevant equations


    3. The attempt at a solution
    I'm pretty sure that to prove that this polynomial is not solvable I am too show that it has exactly 3 roots. That means that it will have 3 roots and 2 complex roots, and thus by a lemma we learned in class the Galois group will be isomorphic to ##S_5## which is not solvable.

    I am a bit confused on how to prove it has 3 real roots. I guess first I'll take the derivative.

    f'(x)=5x^4-12x^3+18x^2+36x = x(5x^3-12x^2+18x+36).

    By the way, this question was a question on last years final, so to solve it I only want to use methods that will be available when I am taking my final in a week (I won't have the general solution to cubic polynomials).

    So at this point I know that the derivative will have at least 1 real zero, when x=0.

    Now I'll take a closer look at (5x^3-12x^2+18x+36). This polynomial looks like it may be reducible. I tried to factor it as ##(ax^2+bx+c)(dx+e)## and match coefficients but I could not solve the system of equations, it had 5 variables and 4 equations and I couldn't figure out how to solve it.

    Anyway, any insight into this from ya'll would be great.
     
  2. jcsd
  3. Nov 30, 2016 #2

    BvU

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    Hi Psycho,


    That doesn't look good. In the brackets there should be a quadratic equation... And now it's easy.
    [Edit]Mistake, sorry. Let me look at it a little better...
     
  4. Nov 30, 2016 #3

    lurflurf

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    First don't forget one must always first show the polynomial is irreducible.
    The discriminant is negative so we have 3 real roots.
    One could also conclude this by noting that p(-2)=-59,p(-1)=5,p(0)=-3,p(1)=19.
    Other methods
    Put the quintic into Bring–Jerrard normal form
    check the Cayley's resolvent for rational roots
     
  5. Nov 30, 2016 #4

    epenguin

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    You mean you think you have to show it has three real roots? What is this lemma exactly? Shurely there can be no theorem that says that having three real and two nonreal roots by itself makes a polynomial equation unsolvable by radicals?

    Isn't discriminant negative negative compatible with one or five real roots, so the argument is not complete?
    The second quoted sentence shows there must be at least three real roots, not yet complete...
     
  6. Nov 30, 2016 #5

    lurflurf

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    not by itself you also need the polynomial to be irreducible
    you are right we also need to see as we can that there are two local extrema one at zero and one between -1 and -2

    for a quintic
    D>0
    one or five real roots all different
    D=0
    one or more repeated roots
    D<0
    three real roots all different
    maybe some people call -D the discriminant it would not matter which as long as you stick with one choice, but such a person would say the given polynomial has positive discriminant.
     
    Last edited: Nov 30, 2016
  7. Dec 1, 2016 #6
    Am I supposed to know the discriminant of a quintic polynomial? I was hoping to figure this out by their algebraic or extremely basic analysis methods (like calculus 1 stuff basically). Showing values for p(r) p(s) and p(x) and showing it crosses the x-axis is great but that only shows it has at least 3 roots, as stated above.
     
  8. Dec 1, 2016 #7

    epenguin

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    Sorry for my late night superficial mistake - it is only the case D > 0 that allows two possibilities. Ignore that comment of mine.

    I do not know what you are supposed to know. It looks like you are doing a course that handles some traditional algebra in an untraditional way. It would do no harm to anyone to know and understand the discriminant of any polynomial, and I dare say your course will get around to that.

    I think you could get number of real roots in your polynomial without this by combination of the information in #3, third line, and Descartes' rule (if you are supposed to know that. :oldbiggrin:).

    This sort of cobbling will often but not always give you a conclusion; the way that always gives you a conclusion for number of ornery real roots of polynomials with ornery real coefficients is Sturm's method.
     
    Last edited: Dec 1, 2016
  9. Dec 1, 2016 #8
    I hope that one of the participants in this thread can confirm for me that I understand the meaning of
    "Polynomial P is not solvable by radicals."​
    I think this means that the value of at least one of the roots of P is not expressible in terms of a finite application of a collection of operators (+, -, ×, /, and any of the n-th roots (where n a positive integer) ) , where these operators are applied to integer operands.

    Is this correct?

    Regards,
    Buzz
     
  10. Dec 1, 2016 #9

    Ray Vickson

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    I don't understand your Lemma. Surely the quintic ##f(x) = (x-1)(x-2)(x-3)(x-i)(x+i)## has three real roots and rwo complex roots, but is most definitely solvable by radicals!
     
  11. Dec 1, 2016 #10
    my bad, the polynomial must also be irreducible. I'll do a quick sketch of the proof.

    It's a degree five polynomial, so the first element we adjoin to Q will have degree 5, and so the final splitting field over Q will be divisible by 5 and since this will be a Galois extension we know that the Galois group will also be divisible by 5. Since the Galois group is permutes 5 roots, it will be a subgroup of ##S_5##, and since 5 divides it then by Cauchy's theorem it will have an element of degree 5, this element must be the 5 cycle. The complex conjugate mapping gives a transposition, and thus we have a 5 cycle and transposition in the Galois group and so this Galois group must be ##S_5## and thus it's not solvable.

    I hope I didn't leave anything out but this is the general idea. Sorry, I probably wrote it sloppily earlier.
     
  12. Dec 1, 2016 #11

    quintic polynomials have a discriminant? And so you're saying since it's negative there must be a complex root and it's conjugate. So if the discriminant is positive it has 5 real roots? So if the roots depend on just the discriminant which is either positive or negative and positive means all rela roots and negative means 2 complex roots, does that mean this polynomial can't have more than 2 complex roots?

    Doesn't your argument of One could also conclude this by noting that p(-2)=-59,p(-1)=5,p(0)=-3,p(1)=19. show that it has AT LEAST 3 real roots? Surely there could be more sign changes in the way negative or way positive real numbers?

    I'm reading about the Descartes sign rule, could that a be a proper way to guarantee the correct answer?
     
  13. Dec 1, 2016 #12

    lurflurf

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    Complex roots come in conjugate pairs a quuntil may have 0,2,4 complex and 1,3,5 real
    As I indicated above
    For a quintic with real coefficients
    D>0
    Either one or five real roots all different
    D=0
    at least one multiple root
    D<0
    Exactly three real roots all different
    Your polynomialis is this case
    Some people might work with -D and reverse the above

    I agree it is tedious to calculate the discriminant but it directly answers the question
    I tried Decartes rule you should too, I think it was inconclusive between 3 and 5 real roots which we already knew
    You are right it is obvious there are 3 real roots but less obvious to prove at least for me
    There are various theorems and ad hoc methods, but nothing comes to mind
     
  14. Dec 2, 2016 #13
    how do I calculate the discriminant? To I need to memorize a formula?
     
  15. Dec 2, 2016 #14

    lurflurf

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    I'm sure there is an easy way to do this. I ust can't think of it.I would not try to memorize the discriminant it is long . I just used https://en.wikipedia.org/wiki/Sturm's_theorem , I don't live finding all those remainders but it was shorter than the discriminant calculation. I noticed if we can show the derivative has a complex root the original polynomial does as well. x is a factor of the derivative which i am sure makes it easy. Some we just have a cubic after factoring x. Still not obvious to me
     
  16. Dec 2, 2016 #15

    epenguin

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    Every polynomial (well lets say in one variable) has a discriminant. It is apart from a numerical factor the product of all squared differences between roots. From that fact you can work out lurflurf's statement that negative discriminant implies three real roots. Really more to the point would be to say that it implies just one pair of conjugate nonreal complex roots.

    But that is saying just in terms of roots. Which you don't have, you just have coefficients. From another point of view, the discriminant is the expression you get whose vanishing is the condition for the polynomial and its derivative to have a common root. That is a very easily memorable formula for any degree, which I'm not writing out now because it is a big determinant, but which you can find in any decent semi-advanced algebra textbook. I.e. probably not yours. :oldbiggrin:

    The said decent textbook will go through the argument to show that these two things are the same, except maybe for a numerical factor. I don't think you are for the moment are required to do it this way since I think you can do an easier calculation as I have already indicated.

    (The discriminant is again just ± the remainder you get on dividing the polynomial by its derivative.)
     
    Last edited: Dec 2, 2016
  17. Dec 2, 2016 #16

    epenguin

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    I think Descartes' rule together with your observation p(-2)=-59,p(-1)=5,p(0)=-3,p(1)=19 allow you to get it.
    (When you say the discriminant is 'direct', it's not that it will always give you the answer - it does in this case because it is negative but it would not tell you by itself the number of real roots if it were positive. Then if you go to higher degree I think the discriminant will never tell you the number of roots by itself. So your discriminant is really just an ad hoc method that worked in this case but I think we're lucky and there are other simpler ad hoc arguments that work in this case. :oldsmile: )

    Aren't those Sturm quantities in fact just certain successive minors of the discriminant determinant?
     
  18. Dec 2, 2016 #17
    Hi @PsychonautQQ:

    I appreciate your "liking" my post # 8. I interpret that to mean you are confirming my definition is correct of the concept
    "Polynomial P is not solvable by radicals".​
    I have since noticed that I should have specified that the polynomial P has integer coefficients.

    Now I would like to ask another question, one about the following Bring-Jerrard quintic equation.
    x5 -x +1 = 0 .​
    The attached PNG file shows that this equation has one real root, with an approximate value of r = -1.1673039778. This seems to me to be the simplest possible example of a
    real algebraic (i.e., non-transcendental) number which is also non-radical.. My question is:
    What would a proof that "r is not radical" look like?​
    I have no idea whatever how one would go about proving the non-radicality of just this one example.

    Can you help me with this?

    Regards,
    Buzz
     

    Attached Files:

    Last edited: Dec 2, 2016
  19. Dec 4, 2016 #18

    epenguin

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    Psychonaut, did you get the answer to the number of real/nonreal roots of your first equation?

    I ask because; -
    see my sig;
    you have gone on to ask several further weird and wonderful questions which I could not possibly attempt nor quite understand what the questions even are, and I find it quite remarkable that anyone who does is not on the other hand familiar with Descartes' theorem etc. with which I can answer that part of your question in a trice!
     
  20. Dec 5, 2016 #19
    Yeah man, I got it, thanks, haha. Yeah, I didn't take my earlier math courses seriously, I didn't actually start liking math until abstract algebra so I suck at everything before that.... Anything involving calculus I'm pretty useless at.
     
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