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  "body": "\n## Statement\n\nLet $R$ be a [[commutative ring|0GYGHKX834]], let $f \\in R[x]$ and let $c \\in R$.\nThen $x - c$ divides $f$ in $R[x]$ if and only if $c$ is a [[root|00EHX6FWC0]] of $f$.\n\n## Remarks\n\nThe proof uses the [[remainder theorem|0TEDQH56RW]]: $f = (x - c) q + f(c)$. So $x - c$ divides $f$ exactly when $f(c) = 0$.\nFor example, $1$ is a root of $x^3 - 1$, and $x^3 - 1 = (x - 1)(x^2 + x + 1)$.\nThe theorem gives the bound on the [[number of roots|0KJVQV3E0F]] of a polynomial over a field.\n\n## Explanations\n\n### L1 human\n\nEach root gives a linear factor.\nIf $f(c) = 0$, then you can write $f$ as $(x - c)$ times another polynomial.\nConversely, a factor $x - c$ makes $f(c) = 0$.\n",
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  "name": "Factor theorem",
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