Divide polynomials in Casio 991ES [closed]

Is there a way to divide to polynomials in Casio 991ES ? For example : $x^4 + x + 1$ with $x^2 + x +1$ ?

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Here is the PDF manual of the product in question. I'm not sure if this question is off-topic. I'm thinking of an evaluation-interpolation scheme.. –  user2468 Mar 22 '12 at 20:43
I am more into believing, it is off-topic. I'd vote to close. –  user21436 Mar 22 '12 at 21:00
No idea. So why am I commenting? (1) Wolfram Alpha can do the job, and, as of now, it is cheaper than a Casio; (2) The division process is quite easy, easier than ordinary division of numbers, and you need to know it. –  André Nicolas Mar 22 '12 at 21:53

closed as off topic by Kannappan Sampath, hardmath, Brian M. Scott, Jonas Teuwen, Asaf KaragilaMar 22 '12 at 23:33

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There is something called Kronecker substitution, such as in here PDF slides and this PDF paper.

The idea is that given two polynomials $f,g \in \mathbb{Z}[x],$ we can encode $f,g$ as long integers by evaluating $f,g$ at a sufficiently large integer $v \in \mathbb{Z}.$ If $v$ is selected appropriately, then you can multiply $f(v)g(v)$ and read off the coefficients of $h(x) \in \mathbb{Z}[x]$ from the digits of $h(v) \in \mathbb{Z}.$

Example (from the PDF slides linked above): Let $f = 41x^3 + 49x^2 + 38x + 29,$ and $g = 19x^3 + 23x^2 + 46x + 21$ be two polynomials in $\mathbb{Z}[x].$ Pick $v = 10^4.$ To get the product $h = fg,$ compute: $$f(10^4) = 41004900380029 \\ g(10^4) = 19002300460021 \\ f(10^4)g(10^4) = \underbrace{0779}_{h_6}\ \underbrace{1874}_{h_5}\ \underbrace{3735}_{h_4}\ \underbrace{4540}_{h_3}\ \underbrace{3444}_{h_2}\ \underbrace{2132}_{h_1}\ \underbrace{0609}_{h_0}$$ Every $4$ digits of the product give a coefficient of $h = fg.$ i.e., $$h = 779x^6+ 1874x^5+ 3735x^4+ 4540x^3+ 3444x^2+ 2132x + 609$$

I don't have a reference on my hand confirming this will work for division. But I tried 1 (one!) example, and it did. I conjecture it works.

Back to OP's question: if the calculator can perform big integer arithmetic, then you can definitely use this "Kronecker trick" to perform polynomial arithmetic.

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