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Homework Help: Roots of cubics

  1. Jan 9, 2010 #1

    Mentallic

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    1. The problem statement, all variables and given/known data
    Not so much a homework problem, but a problem that is annoying me because of its simplicity.

    Not all cubic polynomials with rational coefficients can be factorized by the rational root theorem (or is this false?). What I am finding hard to comprehend is how a cubic with only irrational or imaginary roots can have all rational coefficients.
    What I'm looking for is an example or two illustrating a completely factorized cubic with non-rational roots that, if expanded, has all rational coefficients.

    3. The attempt at a solution
    In an attempt to figure out how it can be done myself I have realized that it is impossible to have one quadratic factor (rational coefficients) that does not have rational roots and the last factor still satisfying this.

    e.g. [tex](x^2+a)(x+b)[/tex] given a>0, the quadratic factor will have imaginary coefficients and of course b will need to be irrational or imaginary, but then if we expand this, the constant will be [itex]ab[/itex] which - if a is rational - will not satisfy the criteria, or if a is such a number that ab becomes rational then we will end up breaking the criteria again since the coefficients of [itex]x[/itex] and [itex]x^2[/itex] will be b and a respectively (which we've already shown have to both be non-rational).

    I've tried a few others, but I'm sure you get the point :smile:

    And of course this problem of mine applies to all polynomials of odd degrees.

    EDIT: Oh and please exclude the obvious (for me anyway) cases of [itex]x^3+a=0[/itex] where a is rational and [itex]a^{1/3}[/itex] is not rational.
     
  2. jcsd
  3. Jan 9, 2010 #2

    Dick

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    I'm not sure why you are excluding x^3+a=0, but ok, this particular example came up in a problem earlier. 8x^3-6x-1=0 has three real roots. None of them are rational. By the rational roots test. The roots are cos(pi/9), and (1/2)*(-cos(pi/9)+/-sqrt(3)*sin(pi/9)). Kind of complicated, but what did you expect? I still like x^3-2=0 much better.
     
    Last edited: Jan 10, 2010
  4. Jan 10, 2010 #3

    Mentallic

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    I excluded that obvious type because I know there were more cases where the coefficients of x and x2 weren't zero, and that one is really simple of course.
    Thanks, I see where they are coming from now.
     
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