$p(x)$ divided by $x-c$ has remainder $p(c)$? [Polynomial Remainder Theorem]

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.

Solution 1:

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 [email protected] 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 $

It is helpful to be aware of the following simple equivalences.

Theorem $\ $ TFAE for a polynomial $\,f\in R[x],\,$ and $\,a\in R\,$ a commutative ring.

$(0)\ \ \ f = (x\!-\!a)q + r\ $ for some $\,q\in R[x],\ r\in R\ \ \ $ [Monic Linear Division Algorithm]

$(1)\ \ \ f\bmod x\!-\!a = f(a)\ \ \ \ \ \ \ $ [Remainder Theorem]

$(2)\ \ \ f(a) = 0\,\Rightarrow\, x\!-\!a\mid f\ \ \ \ $ [Factor Theorem]

Proof $\ (0\Rightarrow 1)\ \ \ f = (x\!-\!a)q + r\,\overset{\large x\,=\,a}\Longrightarrow\, r=f(a)\,\Rightarrow\,f\bmod x\!-\!a = r = f(a) $

$(1\Rightarrow 2)\ \ \ f\bmod x\!-\!a = f(a) = 0\,\Rightarrow\, x\!-\!a\mid f$

$(2\Rightarrow 0)\ \ \ g := f-f(a)\,$ has $\,g(a) = 0\ $ so $\ g = f-f(a) = (x\!-\!a)q$

Remark $ $ The polynomial division algorithm always works in any polynomial ring for divisors that are monic (lead coef $= 1$ or a unit) such as $\,x-a\,$ above, see here.

Solution 2:

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?