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Homework Help: Cubic Polynomials

  1. Jan 11, 2010 #1
    1. The problem statement, all variables and given/known data
    I think I saw another thread answer this question, but I was a little lost whilst reading it.
    I have just recently learned of the rational root theorem and was using it quite happily; figuring out what possibly answers went with cubic and quartic polynomials gave new meaning to guess and check.
    But then I realized something strange, I am aware that, because of conjugates, complex and imaginary solutions to polynomials come in pairs. I am also aware that the amount of roots in a polynomial stem from its degree. This became confusing to me when I came across an equation that was not solved using one of the roots found with the rational root theorem.
    I am confused on how it is possible for a cubic polynomial to have no rational roots, and thus, three imaginary or complex roots - even though complex and imaginary numbers must come in pairs.
    I asked my teacher, but he did not know.


    2. Relevant equations

    This was the equation that spurred the whole confusion:
    2x[tex]^{3}[/tex]-5x[tex]^{2}[/tex]-9x+13=0

    this is a different one my teacher showed the class after I asked him about it:
    2x[tex]^{3}[/tex]-9x[tex]^{2}[/tex]-11x+8=0


    3. The attempt at a solution

    I thought it was as simple as a multiplicity. But that wouldn't work. I am really at a lost here.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 11, 2010 #2

    rock.freak667

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    You are confusing rational with real. Rational and irrational numbers are both real numbers. So you can have a cubic such as (x-√2)(x-√3)(x-√5)=0, the roots are not rational but all are real.
     
  4. Jan 11, 2010 #3

    Dick

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    Right. But that also doesn't lead to a polynomial with rational coefficients. How about (x-(-1/2+i*sqrt(3)/2))*(x-(-1/2-i*sqrt(3)/2))*(x-1)=x^3-1? If you want an even more complicated example with three real roots try https://www.physicsforums.com/showthread.php?t=368349
     
  5. Jan 11, 2010 #4

    Dick

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    Ooops. Forget my first example. That does have a rational root. Sorry. If you change it to x^3-2 and scale them all by 2^(1/3), then it should be ok.
     
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