Why is $x^4+x^2+1$ over $_2$ a reducible polynomial? What do I misunderstand?

Question

I don't quite understand when a polynomial is irreducible and when it's not.

Take $x^2 +1$ over $_3$.

As far as I know, I have to do the following:

0 1 2 using $x \in _3$

1 2 2 using $p(x)$

I calculated it like that:

$(0^2 + 1) \mod 3 = 1$

$(1^2 + 1) \mod 3 = 2$

$(2^2 + 1) \mod 3 = 1$

This is irreducible because in none of them the result is $0$.

Now take $x^2 + 1$ over $_2$

The same approach:

0 1 using $x \in _2$

1 2 using $p(x)$

$(0^2 + 1) \mod 2 = 1$

$(1^2 + 1) \mod 2 = 0$

This is reducible because the result is $0$ in the latter case.

Now take $x^4+x^2+1$ over $_2$.

0 1 using $x \in _2$

1 1 using $p(x)$

$(0^4+0^2 + 1) \mod 2 = 1 $

$(1^4+1^2 + 1) \mod 2 = 1 $

Why is this polynomial still reducible even though we get both times $1$ as a result?

Can someone clarify?

Answer 1

Because, just like in almost all fields, it is possible that a polynomial of degree $4$ is the product of two polynomials of degree $2$, neither of which happens to have a root.