The polynomial remainder theorem states that when a polynomial $P(x)$ of degree $> 0$ is divided by $x-r$ ($r$ being some constant) the remainder is equal to $P(r)$, that is:
$$\begin{array}l If & \quad P(x) = (x-r)Q(x)+R \\ then & \quad P(r) = R \end{array}$$ The algebra and the graphic representation make sense; the question is why. Why is the functional relationship between $r$ and $R$ the function $P(x)$? What are the mechanics, so to speak, that produce this result? Or is it just a fortunate algebraic "accident"?

  • $\begingroup$ It's the first time I see someone other than me identifying 'why' not with plots, but with something else, in this case algebra. I salute you. $\endgroup$
    – Git Gud
    Jun 3, 2014 at 22:08

2 Answers 2


Polynomial long division gives a solution of the equation

$$f(x)=q(x)g(x) + r(x)\,,$$

where the degree of $r(x)$, is less than that of $g(x)$.

If we take $g(x) = x-a$ as the divisor, giving the degree of $r(x)$ as $0$, i.e. $r(x) = R$:

$$f(x)=q(x)(x-a) + R\,.$$

Setting $x=a $, we obtain: $f(a)=R.$

  • $\begingroup$ The algebra is a representation of certain logic; the representation is clear, the logic less so. $\endgroup$
    – Docom
    Jun 4, 2014 at 12:38
  • $\begingroup$ @Docom The logic is given in Bill's answer. $\endgroup$
    – Hakim
    Jun 4, 2014 at 12:43
  • $\begingroup$ Indeed. Thanks! $\endgroup$
    – Docom
    Jun 4, 2014 at 12:54

It is clearer in congruence language, $\ {\rm mod}\ \,x\!-\!r\!:\,\ x\equiv r\,\Rightarrow\ P(x)\equiv P(r).\ $ This is true because polynomials are composed of sums and products, and congruences are equivalence relations that are compatible with sums and products, i.e. $\, A\equiv a,\,B\equiv b\,\Rightarrow\, A+B\equiv a+b,\ AB \equiv ab.\,$ See the proofs of the Congruence Sum, Product and Polynomials Rules below (written for the ring of integers $\,\Bbb Z,\:$ but also valid in any polynomial ring (or any commutative ring).

Congruence Sum Rule $\rm\qquad\quad A\equiv a,\quad B\equiv b\ \Rightarrow\ \color{#c0f}{A+B\,\equiv\, a+b}\ \ \ (mod\ m)$

Proof $\rm\ \ m\: |\: A\!-\!a,\ B\!-\!b\ \Rightarrow\ m\ |\ (A\!-\!a) + (B\!-\!b)\ =\ \color{#c0f}{A+B - (a+b)} $

Congruence Product Rule $\rm\quad\ A\equiv a,\ \ and \ \ B\equiv b\ \Rightarrow\ \color{blue}{AB\equiv ab}\ \ \ (mod\ m)$

Proof $\rm\ \ m\: |\: A\!-\!a,\ B\!-\!b\ \Rightarrow\ m\ |\ (A\!-\!a)\ B + a\ (B\!-\!b)\ =\ \color{blue}{AB - ab} $

Congruence Power Rule $\rm\qquad \color{}{A\equiv a}\ \Rightarrow\ \color{#c00}{A^n\equiv a^n}\ \ (mod\ m)$

Proof $\ $ It is true for $\rm\,n=1\,$ and $\rm\,A\equiv a,\ A^n\equiv a^n \Rightarrow\, \color{#c00}{A^{n+1}\equiv a^{n+1}},\,$ by the Product Rule, so the result follows by induction on $\,n.$

Polynomial Congruence Rule $\ $ If $\,f(x)\,$ is polynomial with integer coefficients then $\ A\equiv a\ \Rightarrow\ f(A)\equiv f(a)\,\pmod m.$

Proof $\ $ By induction on $\, n = $ degree $f.\,$ Clear if $\, n = 0.\,$ Else $\,f(x) = f(0) + x\,g(x)\,$ for $\,g(x)\,$ a polynomial with integer coefficients of degree $< n.\,$ By induction $\,g(A)\equiv g(a)\,$ so $\, \color{#0a0}{A g(A)\equiv a g(a)}\,$ by the Product Rule. Hence $\,f(A) = f(0)+\color{#0a0}{Ag(A)}\equiv f(0)+\color{#0a0}{ag(a)} = f(a)\,$ by the Sum Rule.

Beware $ $ that such rules need not hold true for other operations, e.g. the exponential analog of above $\rm A^B\equiv\, a^b$ is not generally true (unless $\rm B = b,\,$ so it follows by applyimg the Polynomial Rule with $\,f(x) = x^{\rm b}).$

  • $\begingroup$ So let me see if I understand you correctly: Since the remainder is just the difference between two numbers, it persists through addition and multiplication operations (eg. If b = a+r, then 5b = 5(a+r)). Since the polynomial is constituted (or constructed) of just addition and multiplication, the remainder persists throughout the series of operations (or transformation) that is the polynomial, or alternatively - one might say the remainder "follows" the divisor throughout the function P(x). So essentially the remainder theorem is just the result of the elementary distributive property. $\endgroup$
    – Docom
    Jun 4, 2014 at 13:21
  • $\begingroup$ @Docom Consider the special case $\,r = 0.\,$ The answer says $\ {\rm mod}\ x\!:\ x\equiv 0\,\Rightarrow\, P(x)\equiv P(0),\,$ because $\,x\equiv 0\,\Rightarrow\,x^n\equiv 0\,$ by the Power Rule (iterated Prodcut Rule), so $\,c_n x^n\equiv c_n 0\equiv 0\,$ for $\,n\ge 1\,$ by the Product Rule, so $\,P(x) = c_0 + c_1 x+\cdots+c_n x^n \equiv c_0 + 0 +\cdots + 0\equiv c_0 = P(0)\,$ by applying the Sum Rule multiple times. $\endgroup$ Jun 4, 2014 at 14:53
  • $\begingroup$ So, mod $\,x,\,$ every polynomial is equivalent to its remainder $= P(0) = c_0 = $ constant term, so arithmetic mod $\,x\,$ is essentially the same as arithmetic of the constant terms. The essence of the matter will become clearer when one studies quotient(residue) rings, and evaluation maps of polynomials rings. Then it boils down to $\,R[x]/(x-r) \cong R.\ \ $ $\endgroup$ Jun 4, 2014 at 14:53
  • $\begingroup$ Abstract algebra it is, then. Thanks. $\endgroup$
    – Docom
    Jun 9, 2014 at 17:04

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