irreducibility of $x^{5}-2$ over $\mathbb{F}_{11}$. I am tasked to show that $x^{5}-2$ is irreducible over $\mathbb{F}_{11}$ the finite field of 11 elements. I've deduced that it has no linear factors by Fermat's little theorem. But showing it has no quadratic factors is proving harder. 
My approach so far is assume it did. Factor the polynomial and remulitply and compare coefficients to obtain a contradiction. I'm having trouble doing that since there are so many cases. I was given then hint 
"How many elements are in a quadratic extension of $\mathbb{F}_{11}$"? The answer is 121 but I don't know how that helps me. Any hints on dealing with the hint would be very nice. Thanks. 
 A: If $x^5 - 2$ has a quadratic factor, it has a root $\alpha$ in a quadratic extension $K$ of $\mathbb{F}_{11}$. Since the group of units of $K$ has $120$ elements, $\alpha^{120} = 1$. Now $\alpha^5 = 2$, so the order of $2$ must divide ...?
A: Absolutely nothing wrong with Daniel's solution. Just offering an alternative route.
We see that $4$ is a primitive fifth root of unity in $\Bbb{F}_{11}$. If $\alpha$
is a zero of $x^5-2$, then you already know that $\alpha$ is in some extension field $K=\Bbb{F}_{11}[\alpha]$ of degree $n>1$. Then the other zeros of the minimal polynomial of $\alpha$ are among $4^k\alpha, k=1,2,3,4$. Therefore there exists an automorphisms $\sigma\in Gal(K/\Bbb{F}_{11})$ such that 
$\sigma(\alpha)= 4^\ell\alpha$, $0<\ell<5$. But because $\sigma(4^\ell)=4^\ell$ this implies that the order of $\sigma$ is a multiple of five. The claim follows.
A: $\, \color{#c00}{a^{\Large 5}\! = 2}\, \Rightarrow\, a^{\large{120}}\! = \dfrac{(\color{#c00}{a^{\Large{5}}})^{\large{25\!\!\!}}}{\color{#c00}{a^{\Large 5}}}\! = \dfrac{(\color{#c00}2^{\Large{5}})^{\large{5\!\!\!}}}{\color{#c00}2} = \dfrac{(-1)^{\Large{5}}\!\!\!}2\ne 1\ $ contra $\ X^{\Large{120}}\! = 1\,$ for $\ 0\ne X\in \Bbb F_{11^{\Large{2}}}\ \ $ QED
