# $p(x)$ 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.

• If $\dfrac{p(x)}{x-c}=q(x)+\dfrac{r}{x-c}$, then $p(x)=(x-c)q(x)+r$. Now let $x=c$... Dec 28, 2011 at 17:08
• Personally I think the following would be a better title, but I’ll leave it up to you: “How to prove polynomial remainder theorem when $c$ is not a root?” and I’d tag it accordingly Dec 6, 2020 at 3:49

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?

• Is it required to say that $F[x]$ is an Euclidean Domain for any field $F$?
– user21436
Dec 28, 2011 at 17:19
• Perhaps you could go on and prove that? Hint : use induction on the degree of the polynomial you divide. Dec 28, 2011 at 17:21
• @KannappanSampath: "Required" by whom? It all depends on what background material you are allowed to assume and what you are not. Dec 28, 2011 at 17:23
• Actually this result is equivalent to the Factor Theorem, since $P(x)-P(c)$ always has $c$ as a root. Anyhow, I agree with you, usually the Factor Theorem is proven by proving this result first. Dec 28, 2011 at 17:44
• @Arturo well I knew this result it is basically what he does in the book but I guess what I was missing was just plug c in for x Dec 28, 2011 at 17:44

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,

• As Bill Dubuque says, division algorithm works too, I just thought I'd be more explicit. Actually this also provides one way to show the Euclidean algorithm works (for the polynomial $(x-c)$ in particular) : write $p(x) = p((x-c) + c)$ and in the expression of $p(x)$ as $\sum_{k=0}^n a_k x^k$, you replace $x^k$ by $(x-c+c)^k$ and expand this by using the binomial theorem and keeping the $(x-c)$ terms together. Dec 28, 2011 at 17:11
• While that works in this special case, it doesn't generalize to higher degree divisors. It's really the division algorithm that is the essence of the matter here, i.e. that $\rm\:F[x]\:$ is a Euclidean domain. More generally, in any polynomial ring the polynomial long division algorithm works for any divisor whose leading coefficient is a unit, i.e. invertible. Dec 28, 2011 at 17:47
• I know it doesn't for higher degree polynomials, but this argument is often forgotten and sometimes more intuitive than the Euclidean algorithm. Being a fan of intuition myself I like to recall this argument every once in a while. Dec 28, 2011 at 18:25
• One can also do this ring theoretically by exploiting the shift automorphism $\rm\ x\to x+c\:,\:$ see my remark in my answer. Dec 28, 2011 at 20:40
• 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. Dec 29, 2011 at 0:55

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 $x-c$, $1$.

• This is essentially a different way of handling the modular equivalences in Bill Dubuque's answer.
– robjohn
Dec 28, 2011 at 19:24
• 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 $\rm\ x\:\equiv\:c\ \Rightarrow\ x^k\:\equiv\:c^k\ \Rightarrow\ \sum a_k x^k\:\equiv \sum a_k c^k\:$ $\:\Rightarrow\:$ $\rm\:p(x)\:\equiv\:p(c)\:.$ Dec 28, 2011 at 20:54

"$p(x)$ 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:

• Yes, I like to show this way too - but it has pedagogical subtleties - see my remark. Dec 28, 2011 at 20:16