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  "body": "\n## Proof\n\nLet $R$ be a [[commutative ring|0GYGHKX834]], let $f \\in R[x]$ with $f \\neq 0$ and let $c \\in R$. For $h \\in R[x]$, $h'$ is the [[formal derivative|0AD4VS02H3]].\n\n1. The formal derivative is additive, and $(r h)' = r h'$ for $r \\in R$. Both follow from the definition, since it acts on each coefficient separately.\n2. We show that $((x - c) g)' = g + (x - c) g'$ for every $g \\in R[x]$. Write $g = \\sum_{k=0}^{n} b_k x^k$.\n3. Then $(x - c) g = \\sum_{k=0}^{n} b_k x^{k+1} - \\sum_{k=0}^{n} c b_k x^k$. By the definition and step 1, its derivative is $\\sum_{k=0}^{n} (k + 1) b_k x^k - \\sum_{k=1}^{n} k c b_k x^{k-1}$.\n4. Also $g' = \\sum_{k=1}^{n} k b_k x^{k-1}$. So $g + (x - c) g' = \\sum_{k=0}^{n} b_k x^k + \\sum_{k=1}^{n} k b_k x^k - \\sum_{k=1}^{n} c k b_k x^{k-1}$.\n5. The first two sums of step 4 add to $\\sum_{k=0}^{n} (k + 1) b_k x^k$, since $(k + 1) b_k = b_k + k b_k$. Also $c (k b_k) = k (c b_k)$. So steps 3 and 4 give the same polynomial.\n6. The [[evaluation homomorphism|0125QVNCFQ]] at $c$ sends $x - c$ to $0$. So step 2 gives $((x - c) g)'(c) = g(c)$ for every $g \\in R[x]$.\n7. Suppose $c$ has [[multiplicity|03XVP8Q4GY]] at least $2$. Then $(x - c)^2$ divides $f$, say $f = (x - c)^2 h$ with $h \\in R[x]$.\n8. Evaluating at $c$ gives $f(c) = 0 \\cdot h(c) = 0$.\n9. Put $g = (x - c) h$, so that $f = (x - c) g$. By step 6, $f'(c) = g(c) = (c - c) h(c) = 0$.\n10. Conversely, suppose $f(c) = 0$ and $f'(c) = 0$. By the [[factor theorem|0K3MWA7DMC]], $f = (x - c) g$ with $g \\in R[x]$.\n11. By step 6, $g(c) = f'(c) = 0$. By the factor theorem again, $g = (x - c) h$ with $h \\in R[x]$.\n12. So $f = (x - c)^2 h$, and $(x - c)^2$ divides $f$. Since the multiplicity is the largest $m$ with $(x - c)^m$ dividing $f$, it is at least $2$.\n\n## Idea\n\nIf $f = (x - c) g$, the product rule gives $f' = g + (x - c) g'$, so $f'(c) = g(c)$. Thus $f'(c) = 0$ says exactly that $c$ is also a [[root|00EHX6FWC0]] of the cofactor $g$, which means that $x - c$ divides $f$ a second time.\n",
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