Let $F=\mathbb{F}_7$. Let $K$ be the splitting field of $x^3+3$ over $F$. Let $L$ be the splitting field of $x^4+4$ over $F$. What is the degree of the compositum $KL$ over $F$?
My ideas so far: Let $f=x^3+3$. Plugging in the elements of $F$ into $f$ shows that $f$ has no linear factors over $F$, hence is is irreducible over $F$. Therefore $[K:F]=3$. Now let $g=x^4+4$. Plugging in the elements of $F$ into $g$ shows $g$ has no linear factors over $F$. How do I show $g$ has no quadratic factors? Maybe it does have quadratic factors, but how do I verify this? Finally, is the degree of the compositum just going to be the product of the degrees of the splitting fields? Or is it the product divided by the degree of the intersection? How would I find the degree of the intersection of $K$ and $L$?
EDIT: $g$ is reducible, indeed into two quadratic factors as pointed out below by J.W. Tanner. So how then can we find the degree of the compositum?