Factorize the polynomial $f(x)=4x^2-4x+8$ over $mathbb{Z}[x]$, $mathbb{Q}[x]$ and $mathbb{Z}_{11}[x]$

Factorize the polynomial $f(x)=4x^2-4x+8$ in $mathbb{Z}[x]$, $mathbb{Q}[x]$ and $mathbb{Z}_{11}[x]$

My approach

First we consider in $mathbb{Z}[x]$.
Let $$f(x)=4x^2-4x+8=2*2*(x^2-x+2)$$ I claim that $2$ and $(x^2-x+2)$ are ireducible,the fisrt is irreduzible since $mathbb{Z}$ is a UFD and PID and then the primes numbers are the irreducible factors, the second comes from , since the discriminant is $Delta<0$ and $(x^2-x+2)$ have degree $2$ then have two complex roots and therefore are irreducible in $mathbb{R}[x]$ therefore since $mathbb{Z}[x]subset mathbb{R}[x]$ then $(x^2-x+2)$ is irreducible and therefore:
$$f(x)=2*2*(x^2-x+2)$$

For the second case in $mathbb{Q}[x]$ is the same argue since $mathbb{Q}[x] subset mathbb{R}[x]$ and $Delta<0$ implies that $(x^2-x+2)$ is irreducible in $mathbb{Q}[x]$ and therefore
$$f(x)=2*2*(x^2-x+2)$$

In the thrid case in $mathbb{Z}_{11}[x]$ notice that there exists $c_1=overline{5}$ and $c_2=overline{7}$ such that $f(c_1)=f(c_2)=0$ and that implies that $c_1,c_2$ are roots of the polynomial and therefore
$$(x)=2*2*(x^2-x+2)=2*2*(x-5)(x-7)$$ and note that
$$(x-5)(x-7)=x^2-12x+35=x^2+10x+2=x^2-x+2$$
Is my approach right?

Someone can give me an other prove of the fact that $x^2-x+1$ are irreducible in $mathbb{Z}[x]$ and $mathbb{Q}[x]$