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  "body": "\n## Proof\n\nLet $R$ be a [[commutative ring|0GYGHKX834]], let $f = a_0 + a_1 x + \\dots + a_n x^n \\in R[x]$ and let $c \\in R$. We first build $q$ directly, then show that every remainder of degree less than $1$ equals $f(c)$.\n\n1. For an integer $k \\ge 1$, let $S_k = \\sum_{j=0}^{k-1} c^{k-1-j} x^j \\in R[x]$.\n2. Since $R[x]$ is commutative, $(x - c) S_k = \\sum_{j=0}^{k-1} c^{k-1-j} x^{j+1} - \\sum_{j=0}^{k-1} c^{k-j} x^j$.\n3. The first sum has the terms $c^{k-m} x^m$ for $1 \\le m \\le k$. The second sum has the terms $c^{k-m} x^m$ for $0 \\le m \\le k - 1$.\n4. All terms with $1 \\le m \\le k - 1$ cancel. So $(x - c) S_k = x^k - c^k$.\n5. By the definition of the [[value|0GJ7T0NK49]], $f(c) = a_0 + a_1 c + \\dots + a_n c^n$. So $f - f(c) = \\sum_{k=1}^{n} a_k (x^k - c^k)$, where the $k = 0$ terms cancel.\n6. By step 4, $f - f(c) = (x - c) q$ with $q = \\sum_{k=1}^{n} a_k S_k \\in R[x]$. So $f = (x - c) q + f(c)$.\n7. The constant polynomial $f(c)$ has [[degree|0DPF6WRZW6]] $0$ or $-\\infty$, so its degree is less than $1$. So step 6 is a division of $f$ by $x - c$ with a remainder of degree less than $1$.\n8. Now let $q' \\in R[x]$ and $r \\in R[x]$ with $f = (x - c) q' + r$ and $\\deg r \u003c 1$. Then $r$ has no nonzero coefficient of degree at least $1$, so $r$ is a constant $r_0 \\in R$.\n9. The [[evaluation homomorphism|0125QVNCFQ]] at $c$ is a ring homomorphism. Applying it gives $f(c) = (c - c) q'(c) + r_0 = r_0$.\n10. So every remainder of the division of $f$ by $x - c$ is the constant $f(c)$, and step 6 gives $q$ with $f = (x - c) q + f(c)$.\n\n## Idea\n\nEach power satisfies $x^k - c^k = (x - c)(x^{k-1} + c x^{k-2} + \\dots + c^{k-1})$. Summing these identities with the coefficients of $f$ shows that $x - c$ divides $f - f(c)$. Evaluating at $c$ then shows that a constant remainder can only be $f(c)$.\n",
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