Regarding the meaning of remainder polynomial

Question

Consider a polynomial $f(x)$. It can expressed in the form $$f(x) = \mu(x) g(x) + r(x)$$ where degree of $f(x) ≥$ degree of $g(x)$. Remainder when $f(x)$ is divided by $g(x)$ is graphically given by the curve joining the points $f(α_1)$, $f(α_2)$, $f(α_3)$,...$f(α_n)$ where $α_1$, $α_2$,...,$α_n$ are the zeroes of $g(x)$.

I was taught this in my high school. I want to mathematical express this fact but I have very little knowledge of inverse functions. Only the thing that I know is if $y = f(x)$, then $x = f^{-1}(y)$. Using this I deduced that if $g(k) =0$, then $k = g^{-1}(0)$.Also, $r(k) = f(k)$. Therefore, the remainder polynomial is mathematical given by $$f(g^{-1}(0))$$ Am I correct in my deduction?

Answer 1

As it is said in the comments we can not use the concept of inverse functions in general. However, we can find coefficients of $r(x)$ using zeroes of $g(x)$.

Suppose $\deg g(x) = n$ then take $r(x) = b_0 + b_1x + \ldots + b_{n-1}x^{n-1}$. If $\alpha_i$ are the roots of $g(x)$ then we see that $f(\alpha_i) = r(\alpha_i)$. So you get following system of linear equation

\begin{align} b_0 + b_1 \alpha_i + \ldots + b_{n-1} \alpha_{i}^{n-1} = f(\alpha_{i}) && i=1,2,\ldots , n \end{align}

Let $D$ be the determinant of the matrix \begin{pmatrix} 1 && \alpha_0 && \alpha_0^2 && \ldots \alpha_0^{n-1} \\ \vdots && \ldots && \ldots && \vdots \\ 1 && \alpha_n && \ldots && \alpha_{n}^{n-1} \end{pmatrix}

And $D_i$ be the determinant of a matrix whose $i^{th}$ column in the above matrix is replaced by $f(\alpha_1) ,f(\alpha_2) , \ldots, f(\alpha_n)$ for $i = 1,2, \ldots , n$

Then by Cramer's rule the coefficient is given by $b_i = \frac{D_i}{D}$