Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
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.
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.
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?
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,
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 $x-c$, $1$.
"$p(x)$ 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:
