First we prove that $P(x) = x^5 - x^4 - 4x^3 + 4x^2 + 2$ is irreducible in $\mathbb{Z}$. Its possible rational roots are $\pm 1, \pm 2$. Since $P(x) = x^2(x-1)(x-2)(x+2)+2$ it is clear that $\pm 2$ and $1$ are not roots and $P(-1) \neq 0$ too. So if $P(x)$ is reducible then it must be written as a product of two nonlinear polynomials. Looking $P(x)$ modulo $2$ we have $P(x) = x^4(x-1)$. Since there is unique factoring in polynomials in $\mathbb{Z}/2\mathbb{Z}$ then if $P(x)$ is reducible then $P(x) = (x^k + 2P_1(x))(x^l(x-1) + 2P_2(x))$, $P_1(x)$ and $P_2(x)$ with integer coefficients and $k, l > 0$. But then $P(0) = 4P_1(x)P_2(x) \iff P_1(x)P_2(x) = \frac{1}{2}$, which is not possible. So $P(x)$ is irreducible.

Let $\alpha$ be a root of $P(x)$. Then $P(x)$ is the minimal polynomial of $\alpha$, and if $\alpha^n = r$, $r$ rational, then $P(x)$ divides $Q(x) = x^n - r$. But looking modulo $2$ again we have $x^5 \equiv x^4 \pmod{P(x)}$, so $x^n \equiv x^4 \pmod{P(x)}$ and so $\alpha$ is a root of $x^4 - r$. This is a contradiction, since the minimal polynomial of $\alpha$ has degree $5$.