Is $x^4+x+1$ irreducible in $\Bbb{Q}[x]$? 
Decide with a proof if $f(x)=x^4+x+1$ is irreducible in $\Bbb{Q}[x]$.

I was thinking of using DeMoivre's Theorem but I'm not sure how.
Thanks!
 A: Much easier than that! Notice that $x^4+x+1 \in \mathbb{Z}[x]$ is primitive. So $x^4+x+1 \in \mathbb{Q}[x]$ is irreducible if and only if $x^4+x+1 \in \mathbb{Z}[x]$ is irreducible. 
We can show that it is irreducible by looking at the $\mod 2$ reduction and showing that for $0,1$, $f(x)=x^4+x+1 \mod 2$ is not $0$. That shows that $f(x)$ has no linear factors. 
Now we show it has no quadratic factors. Suppose it was reducible. Then it has to be the product of $2$ quadratic factors. But there are only $3$ reducible polynomials in $\mathbb{Z}_2[x]$ so only $1$ reducible one. That is $x^2+x+1$. But the square of this polynomial in $\mathbb{Z}_2[x]$ is not $f(x)$. So $f(x) \in \mathbb{Z}_2[x]$ is not reducible and is then irreducible in $\mathbb{Z}[x]$ and hence also $\mathbb{Q}[x]$. 
A: Let $f(x)=x^4+x+1$. Note that $f(x)$ has no rational root. Hence there will no linear factor of $f(x)$ and hence there will be no cubic factors also. So if $f(x)$ is reducible then there will only quadratic factor only. Clearly this will be of the form $(x^2+ax+1)(x^2+bx+1)$ which is as $x^4+(a+b)x^3+(2+ab)x^2+(a+b)x+1 $. Equating coefficients of similar terms we then get $a+b=0, ab=-2, a+b=1$. In otherwords, $0=1$, contradiction. Hence the proof
A: The $f(2)=19$ is a prime, and therefore $f(x)$ is irreducible in $\mathbb{Z}[x]$ by Cohn's irreducibility criterion, and especially it is irreducible in $\mathbb{Q}[x]$.
