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

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: $f(a)=R.$
