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Factorization of polynomials

  1. Aug 20, 2011 #1
    1. The problem statement, all variables and given/known data

    determine whether the following polynomials are irreducible over Q,

    i)f(x) = [itex]x^5+25x^4+15x^2+20 [/itex]
    ii)f(x) = [itex]x^3+2x^2+3x+5 [/itex]
    iii)f(x) = [itex]x^3+4x^2+3x+2 [/itex]
    iv)f(x) = [itex]x^4+x^3+x^2+x+1 [/itex]


    2. Relevant equations



    3. The attempt at a solution

    By eisensteins criterion let f(x) = [itex]a_n x^n+a_{n-1} x^{n-1}+...a_0[/itex]
    if there exists p, a prime such that p does not divide [itex]a_n[/itex] , p divides [itex]a_{n-1}[/itex],...,p divides [itex]a_0[/itex] and [itex]p^2[/itex] does not divide [itex]a_0[/itex] then f(x) is irreducible over Q

    So i) if p=5 => it is irreducible over Q

    but not sure how to go about the others...
     
  2. jcsd
  3. Aug 20, 2011 #2
    I think ii) and iii) are both reducible but iv) is irreducible as4-1 =3, a prime
     
  4. Aug 20, 2011 #3

    HallsofIvy

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    Since ii and iii are cubics, if they were reducible, they would have to have at least one linear factor, so at least one rational root. By the "rational root theorem", any rational root to ii would have to be [itex]\pm 1[/itex] or [itex]\pm 5[/itex]. Check whether any of those is a root. Similarly, any rational root to iii would have to be [itex]\pm 1[/itex] or [itex]\pm 2[/itex].

    (Clearly neither ii nor iii has a positive root so you really only have two values to check in each problem.)
     
  5. Aug 20, 2011 #4

    micromass

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    For (iv), the roots are easy to compute since

    [tex](x^5-1)=(x-1)(x^4+x^3+x^2+x+1)[/tex]
     
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