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Math Help - Polynomials are irreducible

  1. #1
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    Polynomials are irreducible

    Use Eisenstein's criterion to show taht each of these polynomials is irreducible over the field of rational numbers. (You may need to make a substitution).

    a) x^4 - 12x^2 + 18x - 24
    b) 4x^3 - 15x^2 + 60x + 180
    c) 2x^10 - 25x^3 + 10x^2 - 30
    d) x^2 + 2x - 5 (substitute x-1 or x+1)

    Here is what I have so far:
    a) p = 2 or 3
    when p = 2, then -24 ≡ 0 mod 2^2
    and when p = 3, then -24 ≡ 3 mod 3^2
    so, p = 3
    also, 1 ≡ 0 mod 3 is not true

    b) p = 3 or 5
    when p = 3, then 180 ≡ 0 mod 3^2
    and when p = 5, then 180 ≡ 5 mod 5^2
    so, p = 5
    also, 4 ≡ 0 mod 5 is not true

    c) p = 5
    so, -30 ≡ 6 mod 5^2
    also, 2 ≡ 0 mod 5 is not true

    d) I am not sure

    Does what I have done look correct? Any help for part d would be greatly appreciated.
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  2. #2
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    Quote Originally Posted by page929 View Post
    Use Eisenstein's criterion to show taht each of these polynomials is irreducible over the field of rational numbers. (You may need to make a substitution).

    a) x^4 - 12x^2 + 18x - 24
    b) 4x^3 - 15x^2 + 60x + 180
    c) 2x^10 - 25x^3 + 10x^2 - 30
    d) x^2 + 2x - 5 (substitute x-1 or x+1)

    Here is what I have so far:
    a) p = 2 or 3
    when p = 2, then -24 ≡ 0 mod 2^2
    and when p = 3, then -24 ≡ 3 mod 3^2
    so, p = 3
    also, 1 ≡ 0 mod 3 is not true

    b) p = 3 or 5
    when p = 3, then 180 ≡ 0 mod 3^2
    and when p = 5, then 180 ≡ 5 mod 5^2
    so, p = 5
    also, 4 ≡ 0 mod 5 is not true

    c) p = 5
    so, -30 ≡ 6 mod 5^2
    also, 2 ≡ 0 mod 5 is not true

    d) I am not sure

    Does what I have done look correct? Any help for part d would be greatly appreciated.
    For a) p=2 is not allowed becuase 2^2=4 \text{ and } 4|24

    However p=3 does do the job because it divides all of the coefficients and 3^3=9 \implies 9 \not{|}24

    Therefore it is irreducible.

    once again for b) You have used a prime that is not allowed 3^2=9 and 9|180 p=5 works.

    For the last one Since both 2 and 5 are relatively prime the theorem does not apply directly. As suggested by the hint:

    p(x)=x^2+2x-5 \implies p(x+1)=(x+1)^2+2(x+1)-5=x^2+4x-2

    Now if we let p=2 the polynomial p(x+1) is irreducible. So this implies that p(x) is irreducible.

    Lets see why. Suppose that p(x+1) is irreduicble but p(x) is then p(x)=(ax+b)(cx+d) for a,b,c,d \in \mathbb{Q}, but then

    p(x+1)=[a(x+1)+d][c(x+1)+d]=(ax+a+d)(cx+c+d) but now p(x+1) is reducible and this is a contradiction.
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