Why is $X^4+1$ reducible over $\mathbb F_p$ with $p \geq 3,$ prime

I think I know, finally!, what you meant with this exercise. Be sure you can follow and prove all the following:

Claim: $\,x^4+1\in\Bbb F_p[x]\,$ is reducible for any prime $\,p\,$

Proof : For $\,p=2\,$ the claim follows from $\,x^4+1=(x+1)^4\bmod 2\,$ , so let us suppose $\,p\,$ is odd.

Note that for any such prime, $\,p^2-1=0\bmod 8\,$ (why?) , so that the cyclic multiplicative group $\,\Bbb F_{p^2}^*\,$ has a cyclic subgroup of order $\;8\;$ , and from here that you have an element of order $\,8\,$ in $\,\Bbb F_{p^2}^*\;$ (and not merely an element s.t. $\,\alpha^8=1\,$ , which is completely trivial). Since

$$x^8-1=(x^4+1)(x^4-1)$$

over any field, all the roots of the right hand side are contained in $\;\Bbb F_{p^2}^*\;$ , and since this is an extension of degree two over the prime field $\,\Bbb F_p\,$ (and observe that $\,x^4+1\,$ is always a polynomial over this prime field!), it cannot be this polynomial is irreducible over the prime field since then any of its roots would form an extension of $\,\Bbb F_p\,$ of degree $\;4\;$ , which is absurd. $\;\;\;\;\;\;\;\;\;\;\;\;\;\square\;$

We can use three different factorizations:

If $2$ is a square mod $p$. (which occurs for $p=\pm1$ mod $8$)

$x^4+1=x^4+2x^2+1-2x^2=(x^2+1)^2-2x^2$ Assume $q^2=2$ mod $p$. Then $x^4+1=(x^2+1+qx)(x^2+1-qx)$.

If $-2$ is a square mod $8$. (which occurs for $p=1$ or $3$ mod $8$)

$x^4+1=x^4-2x^2+1+2x^2=(x^2-1)^2-(-2)x^2=(x^2-1+rx)(x^2-1-rx)$, where $r^2=-2$ mod $8$.

Finally, if $-1$ is a square mod $8$. (which happens for $p=1$ mod $4$)

$x^4+1=x^4-(-1)=(x^2+r)(x^2-r)$, where $r^2=-1$ mod $p$.

Checking if all the cases are covered.

All (odd) primes are congruent to either $1,3,5,7$ mod $8$. The cases $1$ and $7$ are covered by the first factorization. The case $3$ by the second. The case $5$ by the third, since $p=5$ mod $8$ implies $p=1$ mod $4$.