Can I use Eisenstein's criterion to show $x^{4}-2x^{3}+2x^{2}+x+4$ is reducible over $\mathbb{Q}$?
Can I say that there does not exist a prime that divides both $2$ and $1$? Or is there another way to show it is reducible? Do I just use rational root test?
You can check this yourself. The basic theorem is that, if this factors over the rationals, in fact it factors over the integers.
This is called Gauss's Lemma; one of them, anyway. Writing as $(x^2 + ax + c)(x^2 + bx + d),$ we find $cd = 4,$ so we have some four possibilities to investigate: $$ (x^2 + ax + 1)(x^2 + bx + 4),$$ $$ (x^2 + ax + 2)(x^2 + bx + 2),$$ $$ (x^2 + ax - 1)(x^2 + bx - 4),$$ $$ (x^2 + ax - 2)(x^2 + bx -2),$$ where, in each one, we see if we can find integer values for $a,b$ that make the multiplication come out as your $x^4 - 2 x^3 + 2 x^2 + x + 4. $ It is pretty easy (in each of four) to find the coefficient of $x^3$ as a combination of $a,b,$ it is also easy to find the coefficient of $x.$ If any of the four gives integer values, the final thing is to see how the coefficient of $x^2$ comes out.