I'm posting to check if I did this right.
First approach was to check roots: $f(0) \neq 0$, $f(1) \neq 0$, $f(2) \neq 0$.
Second approach: Brute Force
My options for degree 1 were: $x$, $x+1$, $x+2$
My options for degree 2 were: $x^2$, $x^2 + 1$, $x^2+2$, $x^2+x$, $x^2+x+1$, $x^2+x+2$
I took out $x$, $x^2$, $x^2+x$ because I have a constant at the end, so there is no way these are factors.
After some long divisions, I arrived at the fact that
$\frac{x^4 + x^3 + x + 2}{x^2 +1} = x^2 + x + 2$ with $r = 0$
I concluded that this polynomial is reducible in $F[x]$.
Is this enough?
All you have to do to check that your factorization is correct is multiply it out.
$\begin{align} (x^2 + 1)(x^2 + x + 2) &= x^2(x^2 + x + 2) + (x^2 + x + 2) \\ &= x^4 + x^3 + 2x^2 + x^2 + x + 2 \\&= x^4 + x^3 + 3x^2 + x + 2\\& = x^4 + x^3 + x + 2\end{align} $
Which is indeed what you started with.