Factor $X^4 + 3$ into irreducible factors in $F_7[X]$ I am not quite sure where to start with this problem. I am new to polynomial rings and want to learn how to factor polynomials in polynomial rings made of fields. 

Factor $X^4 + 3$ into irreducible factors in $F_7[X]$.


You are working in $\Bbb F_7[x]$, the ring of polynomials with coefficients in the finite field $\Bbb F_7$. As Will Jagy also suggested, $3\equiv -4$ mod $7$, hence your polynomial is basically $x^4-4$. 
$$x^4-4=(x^2+2)(x^2-2)$$
We have other two factors to check. Notice that $3^2=9\equiv 2$ mod $7$, hence the second factor splits in $(x+3)(x-3)$, that is
$$x^4+3=x^4-4=(x^2-2)(x^2+2)=(x+3)(x-3)(x^2+2).$$
Finally, consider $(x^2+2)$. You can check by hand that no element in $\Bbb F_7$ is a solution of $q(x)=x^2+2$; namely, for any $a\in \Bbb F_7,$ we have that $a^2+2\not\equiv 0$ mod $7$. Thus it is irreducible.
For these reasons, the total factorization of $x^4+3$ in $\Bbb F_7[x]$ is 
$$x^4+3=(x+3)(x-3)(x^2+2)$$
