The relationship between the intercepts and the remainder in the remainder theorem

Question

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"?

Answer 1

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).

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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 $\,f(x) = x^{\rm b}).$

Answer 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: $f(a)=R.$