Is this polynomial irreducible in $\mathbb{Q}$? this is a really easy question but I cant find an answer;
In need to see if $x^4+x^2+x+1$ is an irreducible polynomial over $\mathbb{Q}$
 A: If it is reducible, it is also reducible over $\mathbb Z$ and has a linear or quadratic factor, again with leading coefficient $1$ and with constant term dividing $1$, i.e. equal to $\pm1$. This rules out the linear factor because neither $1$ nor $-1$ is a root. Remains the case
$$ \begin{align}x^4+x^2+x+1&=(x^2+ax\pm1)(x^2+bx\pm1)\\&= x^4+(a+b)x^3+(ab\pm2)x^2\pm(a+b)x+1\end{align}$$
We conclude $a+b=0$ from the coefficient of $x^3$, but $a+b=\pm1$ from the coefficient pf $x$, hence there is no such factorization.,
A: A quick summary: The rational root test eliminates the possibility of linear factors. Modulo two study (see below) then eliminates the possibility of quadratic factors.

More details:
It suffices to study irreducibility in $\Bbb{Z}[x]$. Assume that it factors. Then you can reduce the factors modulo two, and get a factorization in $\Bbb{Z}_2[x]$. But modulo two we have
$$
p(x)=x^4+x^2+x+1=(x+1)(x^3+x^2+1)
$$
where both factors are irreducible in $\Bbb{Z}_2[x]$. This is clear with the linear factor. For its part the cubic factor has no zeros in $\Bbb{Z}_2$, and hence no linear factors either, so it, too is irreducible.
So if $p(x)$ factors in $\Bbb{Z}[x]$, it must be the product of a linear factor and a cubic factor. 
But, as others have pointed out, the rational root test eliminates the possibility of a linear factor, so the only remaining possibility is that $p(x)$ is irreducible in $\Bbb{Z}[x]$.
A: As it's a quartic, it's either a product of:


*

*A linear and a cubic.

*Two quadratics.


The first would imply that this polynomial has a rational root, which it doesn't due to the rational root test. The second case can be approached by brute force (write out the product of two arbitrary quadratics, identify coefficients, and show the resulting system has no rational solutions).
A: Try factor it into a product of two monic  quadratics and try  solve the equations you get for the coefficients. (note that this polynomial doesn't have a root by the rationale root theorm so this is the only possible way to factor it)  
