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A polynomial equation

  1. Dec 26, 2013 #1
    1. The problem statement, all variables and given/known data

    Solve the following equation:

    [tex] x^4 + 12x^3 + 46x^2 + 60x + 20 = 0 [/tex]


    2. Relevant equations

    Well, I know how to solve simpler equations, in which the unknown dosen't appear at a power higher than 3. I tried to factor this polynom but I didin't suceed.


    3. The attempt at a solution
     
  2. jcsd
  3. Dec 26, 2013 #2

    NascentOxygen

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    If you search google for solving a quartic equation, you will see there is a formula available that gives the exact solution. It is more complicated than the formula you have been using for solving a cubic equation.

    Where an approximation (to any desired accuracy) is acceptable, there are any number of iterative methods for solving polynomials of any order.

    If this were an exam question, it is often the case that you are expected to notice one easy solution (such as x=1. or x= -2) so you can factorize the quartic into two factors with one factor now being a cubic, and you say you know how to solve a cubic.

    It is often handy to know how to check your homework at wolframalpha.com http://m.wolframalpha.com/input/?i=x4+12x3+46x2+60x+20=0&x=0&y=0

    http://imageshack.us/scaled/landing/109/holly1756.gif [Broken]
     
    Last edited by a moderator: May 6, 2017
  4. Dec 26, 2013 #3

    Ray Vickson

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    There are some general things you can try first.

    (1) The rational root theorem; see http://en.wikipedia.org/wiki/Rational_root_theorem . In this case you need to look at all the (integer) factors of +20 (which are +-1, +-2, +-4, +-5, +-10, +-20), to see if any of them solve the equation. If one (or more) of these values "works" you will be able to at least partially factor the polynomial, and then need only deal with another lower-degree equation. If none of them "work" the roots are irrational, so factoring is inapplicable, and you would be stuck dealing with a harder problem.

    (2) Descarte's Rule of Signs; see http://en.wikipedia.org/wiki/Descartes'_rule_of_signs . In this case it implies that the equation has no positive roots; that is easy to see directly, since all the terms have positive coefficients, so all the terms in x^4, x^3,..., increase to larger positive values as x > 0 increases, and the terms are all added together with the same signs. If you look at the polynomial for x < 0 (say by setting x = -t with t > 0) you will see that the successive signs are +, -, +, -, +. There are 4 sign changes, so the Rule of Signs implies that there are either 4, 2 or 0 real roots in t > 0.

    Beyond these general aspects there is not a lot you can do with the equation, except to either use the known "exact" formulas for solving a quartic equation, or use a numerical method. In such questions it is always a good idea to plot the polynomial first, to get a rough idea where the roots (if any) will lie.
     
  5. Dec 26, 2013 #4

    Mark44

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    Per the rules here at PF, you must show what you have tried, even it it wasn't successful.
     
  6. Dec 26, 2013 #5
    @Ray Vickson and @NascentOxygen thank you very much for your explanations! It helped a lot, I managed to solve it and I have also learnt how to divide polynomials, and why does it work:).

    @Mark44 I really really respect and appreciate this forum, but I didn't have what to show. I tried to factor that polynomial on a piece of paper, but it yielded nothing. I mean I just tried the standard methods, because I didn't know how to do it properly. I didn't know about the divison of polynomial, and stuff like that. If I had written: "I tried to factor the equation but if failed" , it would have been ok ? Because now I don't have something else to write there.
     
  7. Dec 26, 2013 #6

    Dick

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    I'd be interested in how you solved it by factoring or polynomial division. I did it by using the first step of solving a general quartic, which is doing a simple change of variables to get rid of the cubic term. When you do that a very lucky thing happens and the linear term goes away too. That makes it easy.
     
    Last edited: Dec 26, 2013
  8. Dec 26, 2013 #7

    Mark44

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    No, that wouldn't have been OK, but if you had shown that a few of the potential roots (as given by the rational root theorem) weren't actually roots, I would have called that a sufficient attempt.
     
  9. Dec 27, 2013 #8
    Well, I didn't solved it by factoring or polynomial division. I have solved as you did, by getting rid of the cubic term. But, my journey in learning how to do it and why it works, has gotten me through polynomial division. I stumbled in a quartic equation, but then I realised that I also don't know how to solve a cubic equation very well. So I did a little digging and I understood the method. :D

    I had not known the rational root theorem before I posted here. Please, pardon me. I really didn't know what to write at that step.
     
  10. Dec 27, 2013 #9

    SteamKing

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    The form of the original equation pretty much eliminates any positive, real solutions. The only remaining candidate solutions will come either from the negative reals or be complex.
     
  11. Dec 27, 2013 #10

    haruspex

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    Having exhausted rational roots, there are a couple of other tricks you can try: a product of simpler polynomials with integer coefficients, or a composition of such polynomials.
    In the present case, it helps that x4 has a coefficient of 1.

    For the product (quadratics, obviously) the general possibility is
    (x2+ax+b)(x2+cx+d)
    Multiplying that out and matching up coefficients gives four equations. You can assume b > d, say, which gives d = 1, 2 or 4, and so on. But there is no solution down this path.

    For the composition of quadratics, we have:
    (x2+ax+b)2+c(x2+ax+b)+d
    This is more successful here.
     
  12. Dec 27, 2013 #11

    Dick

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    Or you could do what DorelX finally did. You can substitute (y-3) for x, the transform that will eliminate the cubic term and discover it will also get rid of the linear term.
     
  13. Dec 27, 2013 #12

    haruspex

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    Sure, but that was rather serendipitous. It is a special case of the composition analysis, in which the inner quadratic is of the form (x+a)2. My point is that it can be generalized usefully.
     
  14. Dec 27, 2013 #13

    Dick

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    No, it's actually not more general. It just looks like it is. If you complete the square on the quadratic in the 'composition analysis', it's exactly the same thing. If you eliminate the cubic term in that form then the linear term will automatically vanish. There aren't that many ways to easily solve a quartic.
     
    Last edited: Dec 27, 2013
  15. Dec 27, 2013 #14

    haruspex

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    Ah, yes. Thanks.
     
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