$ x^4 +x+1$ in $\mathbb{Z}/2\mathbb{Z}[x]$ is an irreducible polynomial.
So far we have only treated quadratic and cubic polynomials, which are irreducible if they do not have any zeros. However, now I want to show that $x^4 + x+1$ in $\mathbb{Z}/2\mathbb{Z}[x]$ is irreducible, I cannot go about checking if it has any zeros, this does not guarantee irreducibility. Is there any clever approach or do I need to determine all the polynomials of lower degree that are irreducible and show that upon division there is always a remainder? $$ \{x, x+1 ,x^2+x+1, x^3 +x+1, x^3 +x^2+1\} $$ are the polynomials I immediately thought of.
You can check easily, that your polynomial has no roots in $\mathbb{Z}/2\mathbb{Z}[X]$ by setting $X=0,1$.
Since it has no roots it has to be $X^4+X+1=(X^2+aX+b)(X^2+cX+d)$ now go ahead and compare the coefficients and deduce a contradiction. Which means that one of a,b,c,d is not an element of $\mathbb{Z}/2\mathbb{Z}$
Edit: To be more specific a contradiction arises, because one of the equalities you obtain does not hold.