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    Can someone please show how to prove it is irreducible.

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    If it isn't irreducible, it must contain a linear or quadratic factor.
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    (Original post by constellarknight)
    If it isn't irreducible, it must contain a linear or quadratic factor.
    I know but how do you prove that it doesn't?

    I know that we can simply find the roots of the polynomial and say that they are not in Q, because this is a quartic.
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    Let the polynomial equal (a+bx+x^2)(c+dx+x^2) and deduce a contradiction. But actually, there's a much faster way: 2 divides -10, but does not divide 1, and 2^2 does not divide 1, so by Eisenstein's criterion, the polynomial is immediately irreducible over the rational numbers.
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    (Original post by constellarknight)
    Let the polynomial equal (a+bx+x^2)(c+dx+x^2) and deduce a contradiction. But actually, there's a much faster way: 2 divides -10, but does not divide 1, and 2^2 does not divide 1, so by Eisenstein's criterion, the polynomial is immediately irreducible over the rational numbers.
    Please can we not use any criterion because we have not come across that so we it is useless to us at the moment...
    Is this OK?

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    Yes, that's fine. Eisenstein's is faster though.
    From Wikipedia:
    To prove the validity of the criterion, suppose Q satisfies the criterion for the prime number p, but that it is nevertheless reducible in Q[x], from which we wish to obtain a contradiction. From Gauss' lemma it follows that Q is reducible in Z[x] as well, and in fact can be written as the product Q = GH of two non-constant polynomials G, H (in case Q is not primitive, one applies the lemma to the primitive polynomial Q/c (where the integer c is the content of Q) to obtain a decomposition for it, and multiplies c into one of the factors to obtain a decomposition for Q). Now reduce Q = GH modulo p to obtain a decomposition in (Z/pZ)[x]. But by hypothesis this reduction for Q leaves its leading term, of the form axn for a non-zero constant aZ/pZ, as the only nonzero term. But then necessarily the reductions modulo p of G and H also make all non-leading terms vanish (and cannot make their leading terms vanish), since no other decompositions of axn are possible in (Z/pZ)[x], which is a unique factorization domain. In particular the constant terms of G and H vanish in the reduction, so they are divisible by p, but then the constant term of Q, which is their product, is divisible by p2, contrary to the hypothesis, and one has a contradiction.
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    (Original post by constellarknight)
    Yes, that's fine. Eisenstein's is faster though.
    From Wikipedia:
    To prove the validity of the criterion, suppose Q satisfies the criterion for the prime number p, but that it is nevertheless reducible in Q[x], from which we wish to obtain a contradiction. From Gauss' lemma it follows that Q is reducible in Z[x] as well, and in fact can be written as the product Q = GH of two non-constant polynomials G, H (in case Q is not primitive, one applies the lemma to the primitive polynomial Q/c (where the integer c is the content of Q) to obtain a decomposition for it, and multiplies c into one of the factors to obtain a decomposition for Q). Now reduce Q = GH modulo p to obtain a decomposition in (Z/pZ)[x]. But by hypothesis this reduction for Q leaves its leading term, of the form axn for a non-zero constant aZ/pZ, as the only nonzero term. But then necessarily the reductions modulo p of G and H also make all non-leading terms vanish (and cannot make their leading terms vanish), since no other decompositions of axn are possible in (Z/pZ)[x], which is a unique factorization domain. In particular the constant terms of G and H vanish in the reduction, so they are divisible by p, but then the constant term of Q, which is their product, is divisible by p2, contrary to the hypothesis, and one has a contradiction.
    I see. Hopefully we will come to this in lectures soon.

    Can I ask, the factorisation I got, I got it off wolfram alpha, how would you come to that factorisation?
    When i solve it for x^2, I get x^2 = 5 +/- i
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    Then set (a+bi)^2=5+i (or 5-i) and solve for a and b.
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    (Original post by cooldudeman)
    Please can we not use any criterion because we have not come across that so we it is useless to us at the moment...
    Have you come across Gauss's lemma yet?
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    (Original post by Gregorius)
    Have you come across Gauss's lemma yet?
    Nope I dont think we will. I don't know why but our lecturer seems to want us to do everything from scratch, well at the moment anyway...

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