If $F$ is a field and $f(x) \in F[x]$, then $f(x)$ has no repeated roots if and only if $(f, f') = 1$, where $f'$ denotes the derivative of $f$.
2026-04-04 03:50:57.1775274657
How do I prove the following theorem?
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Hint: Suppose $f(x) = (x- \alpha)^m g(x), m \geq 2$ then $f'(\alpha) = 0.$ On the other hand, let $f(x) = (x - \alpha)h(x)$ and $f'(\alpha) = 0.$ Then show that $f(x) = (x - \alpha)^2h_1(x).$