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How do I show that $x^2+8x-2$ is irreducible over $\mathbb{Q}$?

I have learnt the basic theory dealing with irreducibility of polynomials belonging to a polynomial field. So, if the given polynomial is to be irreducible, I should be able to find two polynomials $h(x),g(x) \in \mathbb{Q}[x]$ such that $f(x)=h(x)g(x)$ and at least one of them is a constant (unit) polynomial.

I tried by breaking the given equation into factors. I get polynomials with irrational co-efficient. Does that imply that the polynomials are irreducible as the coefficients must be rationals?

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    $\begingroup$ In this case, since it's a quadratic, if it were reducible, you could factor it into two linear polynomials with rational coefficients. That would mean it has a rational zero (or two, but they might coincide). $\endgroup$ – Daniel Fischer Sep 20 '13 at 15:10
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    $\begingroup$ @Aman Mittal Use Eisenstein Criterium with $p=2$ $\endgroup$ – bateman Sep 20 '13 at 15:10
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    $\begingroup$ @DanielFischer ok so my understanding is right. does that imply that over $R$ this same polynomial becomes reducible ? $\endgroup$ – Aman Mittal Sep 20 '13 at 15:12
  • $\begingroup$ Yes, it has two real zeros, so it's reducible over $\mathbb{R}$. $\endgroup$ – Daniel Fischer Sep 20 '13 at 15:13
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One word : Eisenstein. Look it up here for instance. Eisenstein's criterion is probably the most well known irreducibility criterion out there.

At first I thought it was $x^n - 8x + 2$ and I didn't look at $n$ (Eisenstein's criterion works for all $n$). Since $n=2$, you can use the quadratic formula for the roots. If one of the root is in $\mathbb Q$, then the discriminant is the square of a rational, hence the other root will also be rational and you will have a factorization over $\mathbb Q$. Otherwise you won't. (In this case both roots are irrational.)

Hope that helps,

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    $\begingroup$ That was very helpful, thanks. $\endgroup$ – Aman Mittal Sep 20 '13 at 15:14
  • $\begingroup$ BTW is there a systematic method to resolve a polynomial into factors like say $x^4+4$ over $Z_5$ $\endgroup$ – Aman Mittal Sep 20 '13 at 15:18
  • $\begingroup$ @AmanMittal : No prob, feel free to ask more questions. Also, if one answer is your favorite/best, feel free to "check" it (right below the downvote button in each answer, there is a check you can turn green to accept an answer). That gives 25 reputation to the user and 2 to you. $\endgroup$ – Patrick Da Silva Sep 20 '13 at 15:18
  • $\begingroup$ @Aman : For the one you wrote, you can write $(x^2)^2 + 4$ and use the quadratic formula to solve for $x^2$, then get two quadratics you can solve using the quadratic formula. However, the general technique to factor a quartic is not very fun to use and is very complicated (I used it a few times, I can tell). Same thing for cubics. It's usually better if you can find the roots by "looking at it" (i.e. in this case it's pretty clear that $\sqrt 2$ times an $8^{\text{th}}$ primitive root of unity will work (a solution to $x^4 = -1$)). $\endgroup$ – Patrick Da Silva Sep 20 '13 at 15:21
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    $\begingroup$ If you want to know if $x^3 - 9$ is reducible $\pmod {11}$ by finding a root that is $4 \pmod {11}$, just compute $4^3 - 9$ and see if it is 0 $\pmod {11}$. In this case it's $55$, so yes. $\endgroup$ – Patrick Da Silva Sep 20 '13 at 15:43
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Hint

Use the Rational root theorem.

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  • $\begingroup$ Considering we can compute the roots for quadratics in a straightforward manner, that's a bit sad... I would use this idea for cubics or higher degree though. $\endgroup$ – Patrick Da Silva Sep 20 '13 at 15:27
  • $\begingroup$ For higher degree than $3$ the rational root theorem only proves there is no linear factor. A quartic can also factor as the product of two irreducible quadratics, for example. $\endgroup$ – Mark Bennet Sep 20 '13 at 15:31
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Another option in addition to all the fine answers on here: in the case of a polynomial in $\mathbb{Z}[x]$ of degree $\leq 3$, if there are no roots mod some integer, then it is irreducible over $\mathbb{Z}$, and then by Gauss's lemma, it is irreducible over $\mathbb{Q}$.

I get that this polynomial has no roots mod 3.

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Hint.

  1. You can use Eisenstein criterion, and Gauss Lemma. or
  2. A quadratic is irreducible over a Field, means it has not linear factors, i.e., it has no roots in that field. ( Use Rational Root Theorem or solve for x)
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