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

Question

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?

Answer 1

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,

Answer 2

Hint

Use the Rational root theorem.

Answer 3

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)

Answer 4

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.