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.


1 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

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

  • $\begingroup$ Yes, I have heard of Lagrange Interpolation before but never really tried to read more about it since method of differences does the job for me everytime. One thing that I'd like to say is that I think the "Horner form" is very similar to the reconstruction formula in method of differences. Nonetheless, thanks for your answer. Sorry for the delay with accepting your answer. I was busy elsewhere. I have accepted the answer now. Thanks again! :D $\endgroup$ May 23, 2015 at 16:59
  • $\begingroup$ @polynomnomnom See Newton interpolation. $\endgroup$ May 23, 2015 at 17:11

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