$x^p-a$ either has a root or is irreducible

Question

My book (A Book of Abstract Algebra, Pinter) is asking me to explain why if $x^p-a$ factors in $F[x]$ then $x^p-a=p(x)f(x)$ where $\text{deg } p,f \le 2$, here $F$ is a field and $p$ is prime.
It seems to me that this question has a typo, since $\text {deg} pf \le \text{deg}p +\text{deg} f\le 4$ so all primes greater than $4$ would not be accounted for in the result. The book makes no further assumptions about the field $F$, and the goal of the exercise is to prove that either $x^p-a$ has a root or it is irreducible. If I assumed that a typo occurred, I would guess that the question would morph into
Explain why if $x^p-a$ factors in $F[x]$ then $x^p-a=p(x)f(x)$ where $\text{deg } p,f \ge 2$, here $F$ is a field and $p$ is prime.
However, this still does not make sense to me because I can think of counterexamples (in both directions of the inequality) where the fact does not hold. Am I misunderstanding the problem or is it ill-stated?

Answer 1

I agree that the problem you're quoting seems non-nonsensical and has counter examples in $\mathbb{R}$. However, the question that you chose for your title is in fact true - it's just a different question than the one you asked. If you're interested in the answer to that problem, see here.