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This is from Pinter, A Book of Abstract Algebra, p.265.

Given $p(x) \in F[x]$ where $F$ is a field, I would like to show that $p(x)$ divided by $x-c$ has remainder $p(c)$.

This is easy if $c$ is a root of $p$, but I don't see how to prove it if $c$ is not a root.

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    If $\dfrac{p(x)}{x-c}=q(x)+\dfrac{r}{x-c}$, then $p(x)=(x-c)q(x)+r$. Now let $x=c$...2011-12-28

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Division Algorithm $\rm\:\Rightarrow\ p(x)\, =\, q(x)\ (x\!-\!c) + r,\ r\in F.\, $ Evaluate at $\rm\ x = c\ \Rightarrow\ r = p(r).\ $ QED

Therefore $\rm\ x - c\ |\ p(x)-p(c)\ $ in $\rm\:R[x],\,$ $\rm \ \forall\ p\in R[x]\:,\ \forall\ rings\ R\quad $ [Factor Theorem]

Said equivalently $\rm\ p(x)\:\equiv\:p(c)\ \pmod{x - c}$

or, in other words, $\rm\ x\,\equiv\,c\ \Rightarrow\ p(x)\:\equiv\: p(c)\,\ $ [also follows by Polynomial Congruence Rule]

E.g. casting nines: mod $\rm 9\!:\ 10\equiv 1\ \Rightarrow\ N= p(10)\equiv p(1) \equiv $ sum of digits of $\rm\, N\,$ in radix $10$.

Note how the result is much clearer in the language of congruence arithmetic.

Remark $\ $ The shift automorphism $\rm\ x\to x+c\ $ reduces the proof of the Factor Theorem to the "obvious" special case $\rm\:c = 0,\:$ e.g. see here and my sci.math post appended below. However, this approach is a bit risky pedagogically since such a proof is not completely rigorous without knowledge that such maps are ring automorphisms. It is essential that students learn how to make rigorous (ring-theoretically!) prior informal arguments about substitution, changing "variables", etc. Much subtlety lies here, e.g. even in the informal notation for polynomials, such as $\rm\: P(X\!+\!c)\:$ below. This automorphism is the essence behind Patrick's answer, and is also implicitly in Pierre's. Of course such shifting is yet another example of transformation-based problem-solving, cf. my recent post on analogously applying a shift so that Eisenstein's irreducibility criterion applies.

asdf wrote to sci.math on 29 Mar 2006 (paraphrased)

How do you prove that for a polynomial P(X)

$\rm P(c)=0\ \Rightarrow\ X-c\ |\ P(X)\ $ i.e. $\rm\ (X-c)\ Q(X)\ =\ P(X)\ $ for some $\rm\ Q(X)\ $

For $\rm\ c=0\ $ it specializes to the obvious case: $\rm\ X\mid P(X) \iff P(0)=0$

If $\rm\ c\ne 0\ $ reduce to $\rm\ c=0\ $ by a shift: $\rm\ X\!-\!c\mid P(X) \iff X\mid P(X\!+\!c) \iff P(c)=0 $

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By the division algorithm, if $a(x)$ and $b(x)$ are any polynomials, and $a(x)\neq 0$, then there exist unique $q(x)$ and $r(x)$ such that $b(x) = q(x)a(x) + r(x),\qquad r(x)=0\text{ or }\deg(r)\lt \deg(a).$

Let $b(x) = p(x)$, and $a(x)=x-c$. Then $r(x)$ must be constant (since it is either zero or of degree strictly smaller than one), so $b(x) = q(x)(x-c) + r.$ Now evaluate at $x=c$.

Note. I find it strange that you say that this is "easy if $c$ is a root of $p(x)$". The Factor Theorem (that $x-c$ divides $p(x)$ when $c$ is a root of $p(x)$) is a corollary of this result. How exactly do you prove it without this?

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    @Kannappan The assumption that the Divison Algorithm holds in $F[x]$ is equivalent to assuming that $F[x]$ is a Euclidean domain. The former simply uses more elementary language, so it will be accessible to readers who have not yet studied the abstract concept of Euclidean domains.2011-12-28
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Here's how it goes : the polynomials $\{1, (x-c), (x-c)^2, \dots \}$ form a basis of the vector space $F[x]$. Write $ p(x) = a_0 + a_1 (x-c) + a_2 (x-c)^2 + \dots + a_n (x-c)^n. $ Then $ p(x) = (x-c) \left( a_1 + a_2(x-c)^2 + \dots + a_n (x-c)^{n-1} \right)+ a_0 $ and you can see that $p(c) = a_0$.

Hope that helps,

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    Of course there are many ways to do it - some correct, some circular. But, alas, I have no clue from what you wrote in your answer which alternative applies. Hence my remarks. Good mathematical exposition does not leave such doubts in the mind of the reader. It would be nice it you elaborated in your answer *precisely* how you propose to deduce said elements form a basis.2011-12-29
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Suppose that $ p(x)=\sum_{k=0}^na_kx^k $ Then $ p(x)-p(c)=\sum_{k=0}^na_k(x^k-c^k) $ For each $k$, we have that $ \frac{x^k-c^k}{x-c}=x^{k-1}+x^{k-2}c+x^{k-3}c^2+\dots+xc^{k-2}+c^{k-1} $ Thus, $ \begin{align} \frac{p(x)-p(c)}{x-c} &=\sum_{k=0}^na_k(x^{k-1}+x^{k-2}c+x^{k-3}c^2+\dots+xc^{k-2}+c^{k-1})\\ &=q(x) \end{align} $ Therefore, $p(x)=q(x)(x-c)+p(c)$, which is another way of writing $p(x)$ divided by $x-c$ leaves a remainder of $p(c)$ since the degree of $p(c)$, $0$, is less than the degree of $x-c$, $1$.

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    Yes, by *linearity*, $\rm\ x-c\ |\ p(x)-p(c)\ $ reduces to the well-known monomial case $\rm\ p(x) = x^k\:.\:$ In terms of congruences $\rm\ x\:\equiv\:c\ \Rightarrow\ x^k\:\equiv\:c^k\ \Rightarrow\ \sum a_k x^k\:\equiv \sum a_k c^k\:$ $\:\Rightarrow\:$ $\rm\:p(x)\:\equiv\:p(c)\:.$2011-12-28
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"$p(x)$ divided by $x-c$ has remainder $p(c)$"

is equivalent to

"$p(x+c)$ divided by $x$ has remainder $p(c)$",

which is equivalent to

"$q(x)$ divided by $x$ has remainder $q(0)$",

which is obvious.

EDIT. This is essentially the argument used By Gauss in Article 43 of the Disquisitiones Arithmeticae. A French translation is available at Internet Archive and at Google Books:

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    Yes, I like to show this way too - but it has pedagogical subtleties - see my remark.2011-12-28