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Multiplicity drop under differentiation (proof of the multiplicity-by-derivatives theorem). Let \(P \in \mathbb{K}[X]\) and \(a \in \mathbb{K}\).
  1. Suppose \(a\) is a root of \(P\) of multiplicity \(m \geq 1\). Show that \(a\) is a root of \(P'\) of multiplicity exactly \(m - 1\). (Hint. Write \(P = (X - a)^m Q\) with \(Q(a) \neq 0\), then differentiate.)
  2. By induction on \(k\), show that for \(k \in \{0, 1, \dots, m\}\), \(a\) is a root of \(P^{(k)}\) of multiplicity exactly \(m - k\).
  3. Deduce: \(a\) is a root of multiplicity exactly \(m\) of \(P\) if and only if $$ P(a) = P'(a) = \dots = P^{(m-1)}(a) = 0 \quad \text{and} \quad P^{(m)}(a) \neq 0. $$

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