= Solution
Interpret the displayed factorization as a proper factorization, with $1\leq r\leq p-1$; that restriction is necessary for the conclusion. In a <splitting field>, the monic degree-$r$ factor has $r$ of the distinct roots of the original <polynomial>. Comparing its coefficient of $X^{r-1}$ gives
$$
-a_1=\sum_{j\in J}(\alpha+j)=r\alpha+\sum_{j\in J}j,
$$
for an $r$-element subset $J\subseteq\mathbb F_p$. Both $a_1$ and the final sum lie in $F$. Since $0<r<p$, the element $r$ is invertible in $F$, so
$$
\boxed{\alpha=r^{-1}\left(-a_1-\sum_{j\in J}j\right)\in F.}
$$
Part (1) then gives complete splitting into distinct linear factors. Thus an <Artin–Schreier polynomial> of this form either is irreducible or splits completely; a trivial factorization with $r=0$ or $r=p$ would not imply a <root> in $F$.
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