The roots of any polynomial over $\mathbb{F}_3$ lie in some finite extension field $\mathbb{F}_{3^n}$.
The roots of $x^4 + 1$ are all primitive eighth roots of unity. Therefore, we know they live in any extension field such that $8 \mid 3^n - 1$.
The smallest such field is the one with $n=2$. However, this is merely a degree 2 extension! The minimal polynomial of an eighth root of unity in characteristic 3, then, is a quadratic polynomial. The irreducible factorization of $x^4 + 1$ over $\mathbb{F}_2$, therefore, must be a product of two quadratic polynomials.
We can say more: if $\zeta$ is a primitive eighth root of unity, then $\zeta$ and $\zeta^3$ are roots of one factor, and $\zeta^5$ and $\zeta^{15} = \zeta^7$ are roots of the other factor. We have a factorization
$$ x^4 + 1 = (x^2 - (\zeta+\zeta^3) x + \zeta^4) (x^2 - (\zeta^5 + \zeta^7) x + \zeta^4) $$
If you construct a realization of the field $\mathbb{F}_9$, you could do the arithmetic to simplify the coefficients to elements of $\mathbb{F}_3$.
But we can cheat: from ordinary arithmetic in characteristic 0, we know the standard eighth roots of unity
- $\zeta = \frac{\sqrt{2}}{2} (1 + \mathbf{i})$
- $\zeta^3= \frac{\sqrt{2}}{2} (-1 + \mathbf{i})$
- $\zeta^5 = \frac{\sqrt{2}}{2} (-1 - \mathbf{i})$
- $\zeta^7 = \frac{\sqrt{2}}{2} (1 - \mathbf{i})$
and so the factorization should simplify to
$$ x^4 + 1 = (x^2 - \frac{\mathbf{i}\sqrt{2}}{2} x + (-1))(x^2 - \frac{-\mathbf{i}\sqrt{2}}{2} + (-1))$$
In characteristic 3, $(\mathbf{i} \sqrt{2})^2 = -2 = 1$, so if we pick $1$ as the product of square roots, we should have
$$ x^4 + 1 = (x^2 - \frac{1}{2} x + (-1))(x^2 - \frac{-1}{2} + (-1))
= (x^2 - 2x - 1)(x^2 + 2x - 1)$$
and if we didn't believe that this "cheat" was actually a valid derivation, we can verify this factorization is correct directly.
You can easily check with the rational root theorem that there are no roots, so we just have to check if there is a factorization into two quadratics. Set $$x^4 - 4x^2 + 9x + 4 = (x^2 + ax + b)(x^2 + cx + d).$$ Comparing coefficients of $x^3$ gives $c = -a$, so we end up with $$x^4 - 4x^2 + 9x + 4 = (x^2 + ax + b)(x^2 - ax + d)$$ where $a,b,d$ satisfy $bd = 4$, $a(d-b) = 9$ and $-a^2 + b + d = 4$. Now the problem is quite easy with some trial and error, for example try all divisors of $4$ for $b,d$.
Best Answer
If we can write $$p(x)= q(x)r(x)$$ say in $\mathbb{Z}$ then we can do that also in $\mathbb{Z}_m$ where $m$ is an arbitrary integer $>1$. So there exists $q_1,r_1$ such that
$$p(x)\equiv_m q_1(x)r_1(x)$$ and $$r(x) \equiv_m r_1(x) \;\;\;\wedge\;\;\; q(x) \equiv_m q_1(x) $$
and we might hope that $r(x) = r_1(x)$ or $q(x) =q_1(x)$.
If we try this in your case for mod $2$ we get $$p(x) \equiv_2 x^4+x \equiv_2 x(x+1)(x^2+x+1)$$
and we see that $x^2+x+1$ actually divide $p(x)$.