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 $\mathbb{Q}[x]$
Your tittle question has the following answer. First, $f(x)=4(x+6)(x+4)$ is the factorisation over the finite field $\Bbb F_{11}$, as you said. Secondly, $x^2-x+2$ is irreducible over $\Bbb Z$ by the rational root theorem, and as a monic polynomial then also over $\Bbb Q$.