# how to do polynomial division

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faa nad el 3 de Nov. de 2011
Respondida: Ahmed J. Abougarair el 18 de Nov. de 2022
hi...
i wanna do polynomial divison given numerator=x^5+x^4+x^3 and
denominator=x^3+x+1 ;remainder should be x...how to implement it in matlab..
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Walter Roberson el 3 de Nov. de 2011
Will the coefficients always be either 0 or 1 ? A binary polynomial?
faa nad el 5 de Nov. de 2011
yes ,it is a binary polynomial

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### Respuestas (4)

Daniel Baboiu el 3 de Nov. de 2011
You have two choices: 1. Use the Symbolic Math Toolbox 2. Store all coefficients as a vector (including the coefficients which are 0), then use this representation to implement division steps as described below: http://en.wikipedia.org/wiki/Polynomial_long_division
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faa nad el 5 de Nov. de 2011
hi..
thanks for ur reply.is there any other options for dividing the polynomial directly ?what we do in mathematics.without extracting the coefficients.
Walter Roberson el 5 de Nov. de 2011
In mathematics, we mentally extract the coefficients in order to do the division.

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Andrei Bobrov el 5 de Nov. de 2011
[a,b]=deconv([1 1 1 0 0 0],[1 0 1 1])
p1=[1 1 1 0 0 0]
p2=[1 0 1 1]
[a b] = deconv(p1,p2)
syms x
k = cellfun(@(y) y*x.^(numel(y)-1:-1:0).',{a b p2},'un',0)
k = [k{:}]
out = k(1) + k(2)/k(3)
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faa nad el 5 de Nov. de 2011
the answer given by u works well..but my aim is to divide two polynomial expressions[1+x+x^2]/[1+x] directly.the output must be polynomial expression

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Walter Roberson el 5 de Nov. de 2011
As you have restricted this to symbolic expressions without ever extracting the coefficients (at least not in code you write, even if it gets done "under the hood"), then the solution is to use the MuPAD Standard Library function pdivide
I could offer a very nice and efficient calculation for polynomials up to order 52 where the coefficients are all 0 or 1, if we are allowed to extract the coefficients in the code (which you could stuff in to a subroutine and never look at again), but I gather that efficiency and simplicity are not important for your purposes.
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Walter Roberson el 6 de Nov. de 2011
Do you have the symbolic toolbox installed and licensed? If not, then you cannot do what you are asking for, as only the symbolic toolbox hides extracting the coefficients of polynomials.
Maria Maximina el 21 de Feb. de 2014
hi! o have one question for you! i know it is long time ago.. but anyway.. jajaja if u do that operation that you suggested, and you get a vector like:
[a,b,polinom]
what do actually a and b mean??? thanks!

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Ahmed J. Abougarair el 18 de Nov. de 2022
syms x y
p = x^3 - x*y^2 + 1;
d = x + y;
[r,q] = polynomialReduce(p,d)
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