
Re: Polynomial (not quite) division
Posted:
Jan 15, 2014 3:22 PM


In article <102d1a32f67d49cdbd17da8f75477404@googlegroups.com>, hockleymitchell@gmail.com wrote:
> .... > (1 + x^2)(x^3 + 1) = x^3 + x + 1 + x^2 > > Now if I were to use the standard polynomial division algorithm (I appreciate > that x^2 + 1 is not a polynomial) to work out > > (x^3 + x + 1 + x^2) / (x^3 + 1) > > then I would obtain a whole part of 1 and a remainder of x + x^2. And this > is clearly not right....
Allowing negative powers actually gives you the quotient 1 + x^2 and remainder 0.
But a simple way to deal with it is to use genuine polynomials. Your x^3 + x + 1 + x^2 = x^2.(x^5 + x^3 + x^2 + 1). Ordinary division of x^5 + x^3 + x^2 + 1 by x^3 + 1 gives x^5 + x^3 + x^2 + 1 = (x^3 + 1)(x^2 + 1) + 0 and then you can just reintroduce the factor x^2.
Ken Pledger.

