= Solution
{c}
The intended hypothesis must be a proper monic factor, with $1\leq r\leq p-1$. Without that restriction the printed deduction is false: every <polynomial> has the displayed trivial factorization with $r=p$, even when it has no <root> in $F$.
By part 1, the roots in a <splitting field> are distinct and have the form $\alpha+j$. The <degree>-$r$ monic factor is the product of $r$ corresponding linear factors. Comparing the coefficient of $X^{r-1}$ using <Vieta's formulas> gives
$$
-a_1=\sum_{j\in J}(\alpha+j)=r\alpha+\sum_{j\in J}j,\qquad |J|=r.
$$
Both $a_1$ and the sum of prime-<field> elements lie in $F$. Since $1\leq r<p$, the scalar $r$ is nonzero and invertible in $F$. Therefore $\alpha=(-a_1-\sum_{j\in J}j)/r\in F$, and part 1 proves \b[the <polynomial> splits into distinct linear factors over $F$]. In particular an <Artin–Schreier polynomial> of prime <degree> is either <irreducible> or splits completely.
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