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Larson 4.2.15

  1. Jan 7, 2008 #1
    1. The problem statement, all variables and given/known data
    Factor x^8+x^4+1 into irreducible factors (i) over the rationals, (ii) over the reals, (iii) over the complex numbers


    2. Relevant equations



    3. The attempt at a solution
    Is there a systematic way to do this? Do I need to just make a good guess of some factor?
     
  2. jcsd
  3. Jan 8, 2008 #2

    morphism

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    Well, you can use some tricks. Usually complex numbers, and in particular roots of unity, are helpful, but I couldn't really see any direct way of using them here. But what I did notice was the following:
    x^8 + x^4 + 1 = x^8 + 2x^4 + 1 - x^4 = (x^4 + 1)^2 - x^4 = (x^4 - x^2 + 1)(x^4 + x^2 + 1)

    And you can factor it further using the same trick. Dunno if this is what you're looking for though. (Oh, and by the way, the factor on the left above is irreducible over Q - it's a cyclotomic polynomial.)
     
  4. Jan 8, 2008 #3

    Gib Z

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    We have at our disposal of a particularly nice trick for this question =] Hint: It's a quadratic in disguise.
     
  5. Jan 8, 2008 #4

    morphism

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    But is that really helpful here? It does give us a factorization over C, but then we still have to work backwards to get factorizations over R and Q, and I don't think this is any easier. Maybe I'm completely missing something though!
     
  6. Jan 8, 2008 #5

    Gib Z

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    Well thats what I was thinking of doing :( Working backwards. With the aid of the conjugate root theorem we know which factors to multiply with each other to get back to the factorization over R, and then with further inspection, back to the rationals.

    Don't know it thats easier, but its an alternative =]
     
  7. Jan 8, 2008 #6
    Well, if we were to work in the forward direction, we could start by first using the rational roots test which tells us that the only only possible rational roots would be +/- 1. Neither of these are roots, and so the polynomial is irreducible as already stated above.

    As for the real numbers, we could make the substitution [itex] u = x^4 [/itex], solve for u, and then solve for x again. However [itex] u^2+u+1 =0 [/itex] has no real roots, and thus certainly x will have no real roots.

    Thus we need only worry about the complex factorization, and so it doesn't matter what's easier.
     
  8. Jan 8, 2008 #7

    morphism

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    Hold on - just because a polynomial doesn't have roots in a field doesn't mean it's irreducible over it. In fact I gave a factorization of x^8 + x^4 + 1 into two quartics with rational coefficients above, and one of these two quartics can be further factored into a pair of quadratics.
     
  9. Jan 9, 2008 #8
    Ah yes, an excellent point. It just means that it doesn't factor into linear terms. However, we can still use this as a tool for figuring out the degree's of any irreducible terms.
     
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