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  "body": "\n## Proof\n\nLet $R$ be a [[commutative ring|0GYGHKX834]], let $f \\in R[x]$ and let $c \\in R$.\n\n1. Suppose that $x - c$ [[divides|04GNK7K4VP]] $f$ in $R[x]$. Then $f = (x - c) h$ for some $h \\in R[x]$.\n2. The [[evaluation homomorphism|0125QVNCFQ]] at $c$ is a ring homomorphism. So $f(c) = (c - c) h(c) = 0 \\cdot h(c) = 0$.\n3. Thus $c$ is a [[root|00EHX6FWC0]] of $f$.\n4. Conversely, suppose that $f(c) = 0$. By the [[remainder theorem|0TEDQH56RW]], there is $q \\in R[x]$ with $f = (x - c) q + f(c)$.\n5. Since $f(c) = 0$, this gives $f = (x - c) q$. So $x - c$ divides $f$ in $R[x]$.\n\n## Idea\n\nThe remainder of $f$ on division by $x - c$ is the number $f(c)$. So $x - c$ goes into $f$ exactly when this remainder is zero, that is, when $c$ is a root.\n",
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