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Factoring polynomial over rings

  1. Nov 28, 2011 #1
    1. The problem statement, all variables and given/known data

    Factor x^3+3x+6 over Z5, Z10, Z, Q and ℝ.

    2. The attempt at a solution

    For Z5, I have the roots of x^3+3x+1 in Z5, x=2=-3 and x=3=-4, so x+3 and x+4 are factors. By long division of x^3+3x+1, it is found that x+3 is a repeated factor, so x^3+3x+6=(x+4)(x+3)^2.

    For Z and Q, Eisenhart's criterion is satisfied with 3, hence x^3+3x+6 is irreducible over Q, and hence Z.

    For Z10 and ℝ... I don't really get how to do this problem. Is there a consistent way to work out the factors for Z10? It seems like a lot of effort. Also, I noticed that x^3+3x+6 has one real root over ℝ, and so I can factor it neatly to a linear factor (with real root) and a quadratic factor with two complex roots, but I only managed to get these roots with a computational method - and they have pretty long but explicit expressions, so again, my question: is there a proper way to work out the roots by hand here?

    Thanks!
     
  2. jcsd
  3. Nov 28, 2011 #2

    I like Serena

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    Welcome to PF, llstelle! :smile:


    First off, x=3 is not a root in Z5, but x=1 is.

    For Z10 you only need 1 root. Is perhaps x=1 or x=2 also a root in Z10?


    For R, I think you're stuck with the complicated solutions.
    But here's a general method that gives the answer fairly easily:


    Starting with [itex]x^3+3x+6=0[/itex].

    Substitute x=y+z, meaning you have a free choice for either y or z.
    So [itex](y+z)^3+3(y+z)+6=0[/itex]

    [itex]\Rightarrow (y^3+z^3+3y^2z+3yz^2) + 3(y+z)+6=0[/itex]

    [itex]\Rightarrow y^3+z^3+3(yz+1)(y+z)+6=0[/itex]


    Choose z such that [itex]yz+1=0[/itex], or [itex]z=-{1 \over y}[/itex]
    Then: [itex]y^3 - {1 \over y^3} + 6 = 0[/itex]
    [itex]\Rightarrow (y^3)^2 + 6(y^3) - 1 = 0[/itex]


    Solve as a quadratic equation and back substituting z gives:
    [tex]x=y+z=\sqrt[3]{-3 + \sqrt{10}} - {1 \over \sqrt[3]{-3 + \sqrt{10}}}[/tex]
    or
    [tex]x=y+z=-\sqrt[3]{3 + \sqrt{10}} + {1 \over \sqrt[3]{3 + \sqrt{10}}}[/tex]

    Note that both solutions are the same root.
     
  4. Nov 28, 2011 #3
    Hello! Oh oops, that was a typo - yes I have x=1 as a root of the equation for Z5. And thanks for the general solution for ℝ! I've never done that before. I've been taught to test numbers until I find a root, then divide the cubic equation by the linear factor to find a quadratic factor, and use the general solution for a quadratic equation for the quadratic factor.. ): This is so much neater!

    OK, I tried each of them for Z10 and I have x=1, 2, 6 or 7 satisfying f(x)=0. If I write (x+9)(x+8)(x+4)(x+3), I get a fourth order polynomial whose fourth order term does not vanish in Z10... How does this happen? Something looks wrong to me.

    (:
     
  5. Nov 28, 2011 #4

    I like Serena

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    :smile:


    A polynomial of degree n can have more than n roots in Zn if n is not prime.
    For instance, [itex]x^2-1[/itex] has 8 roots in Z24: [itex]\pm 1, \pm 5, \pm 7, \pm 11[/itex].
     
    Last edited: Nov 28, 2011
  6. Nov 28, 2011 #5

    Dick

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    And would probably indicate that x^3+3*x+6 doesn't have a unique factorization. Z10 has zero divisors. You can probably find several factorizations.
     
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