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Factoring a polynomial!

  1. Nov 16, 2008 #1
    the question is this: factor the following polynomial with integer coefficients:


    my first thought is that since this polynomial doesn't have real roots such a factorization isn't even possible, but i really don't have any more clever idea how to factor this one.
  2. jcsd
  3. Nov 16, 2008 #2


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    The only possible rational roots of that are 1 and -1. Obviously neither of those satisfies the equation [itex]x^{5024}+ x^{1004}+ 1= 0[/itex] so it has no rational roots and so cannot be factored with integer coefficients.
  4. Nov 16, 2008 #3
    Ok now, say we take out the requirement that the coefficients be integers. Then, if we are simply looking to express that polynomial as a product of two other polinomials, then will it include inevitably complex coefficients? Or, how would that factorization in terms of two other polynomials look like? Since, we won't have real roots, i see that there will only be complex roots, so even the factorization, from my point of view, seems to include inevitably complex coefficients, right?
  5. Nov 16, 2008 #4


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    Not necessarily. Since it has real coefficients, all complex roots must come in "complex conjugate" pairs: a+ bi and a- bi. Any factorization into linear factors would have to be of the form (x-(a+bi))(x-(a-bi)) but those multiplied together: ((x-a)- bi)((x-a)+ bi)= (x-a)2+ b2= x2- 2ax+ a2+ b2 so it can be factored into 5024/2= 2512 quadratic factors with real coefficients.
  6. Nov 19, 2008 #5
    so, does it mean that in general any polynomial, say of the form

    [tex] x^n+x^m+1[/tex] where both n,m are even integers cannot be factored over the integers. That is cannot be written, say as a product of two polynomials with integer coefficients?
  7. Nov 21, 2008 #6
    I say this is possible to be written, right? SInce, i can already find a counterexample if you say otherwise? So, when would it be the case? i assume wehnever n, is an integral multiple of m, right?
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