Generalizing the Remainder Factor Theorem

Question

Today, I spent most of my time developing a systematic procedure for finding remainder polynomial when higher degree polynomials are divided by some polynomial of degree $\leq$ the degree of the dividend polynomial.

My method uses formulating a result based off the division algorithm and then getting the values of $r(x)$ at points which are roots of $d(x)$ where $r(x)$ is remainder polynomial corresponding to divisor $d(x)$.

We get values one more than the degree of $r(x)$, so we use method of differences to construct a difference table for $r(x)$ and then reconstruct $r(x)$, hence obtaining our answer.

I tried this devised method on an example I formulated. The problem is that I have no place to verify whether my final result is correct or not, implying whether my devised method actually works or not.

The example is "Remainder when $x^{10}$ is divided by $(x-1)(x-2)(x-3)$".

I'm getting the answer as $(28501x^2-84480x+55980)$. I'd appreciate if someone can verify whether my answer is correct or not.

P.s - I'd appreciate verification in the form of a comment simply stating whether my answer is correct or not. I don't need any form of hints or solution to the problem itself. Thanks.

Answer 1

It is correct since it takes the values $\,1^{10},2^{10},3^{10}$ at $\,x = 1,2,3.\,$ Generally we can use Lagrange interpolation, wich is a special case of CRT = Chinese Remainder Theorem

$$ f(x)\, =\, f(1) \dfrac{(x-2)(x-3)}{(1-2)(1-3)} + f(2) \dfrac{(x-3)(x-1)}{(2-3)(2-1)} + f(3) \dfrac{(x-1)(x-2)}{(3-1)(3-2)}$$

or in Horner form

$$ f(x)\, =\, f(1) + (x-1)\left[f(2)-f(1) + (x-2)\left(\frac{f(3)+f(1)}2-f(2)\right)\right]$$