factorization of Multivariate polynomial

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Bruno Luong 2023-3-6

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评论： Walter Roberson 2023-12-2

Hello, I want to factorize a multivariate polynomial in a complex field. I just don't understand why the result of this simple polynomial

syms x y

F = factor(x^2+y^2,'FactorMode','complex')

F =

and not (x + i*y)*(x - i*y)?

Is there any third-party toolbox/code that can carry out the problem of factorization multivariate polynomial?

NOTE: in 1D it's equivalent to finding the roots of the polynomial3

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回答（2 个）

Manikanta Aditya 2023-3-17

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Hi Bruno,

As per my understanding, you want to factorize a polynomial in a complex field, and you are getting result of this simple polynomial.

The reason why the factorization of x^2+y^2 using ‘factor’ function in MATLAB returns a different result than (x + i*y)*(x - i*y) is because ‘factor’ function only returns factors with real coefficients by default.

Since the factors (x + i*y) and (x - i*y) have complex coefficients, they are not returned by the ‘factor’ function. If you want to obtain factors with complex coefficients, you can use the factor function with the option 'FactorMode','full'.

syms x y

F = factor(x^2+y^2,'FactorMode','full')

F =

‘Factormode’ parameter set to ‘full’ will include complex conjugate pair of factors.

For further reference, please go through this link for knowing about symbolic factors:

https://www.mathworks.com/help/symbolic/factor.html

I hope this resolves the issue you are facing.

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Paul 2023-3-23

编辑：Paul 2023-3-23

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syms x y;
p = x^2 + y^2;

The following line isn't actually a documented syntax of solve. There is a note in the inputs section that says "By default, solve uses the variable determined by symvar." I think that should really be symvar(eqn,1). Contrast with diff and the very first syntax and its description.

roots_p = solve(p);

Anyway, this line actually soved for symvar(p,1), so the correct expression would be

factors_p = prod(symvar(p,1) - roots_p)

factors_p =

Bruno Luong 2023-3-23

@Paul, this is similar to @John D'Errico solution but without doing explicit division. This is neat, thanks.

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John D'Errico 2023-3-17

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This is a bit of a hack of course, since factor seemingly should have solved the problem with the appropriate setting. The problem with the complex FactorMode is the polynomial is not one with constant coefficients.

syms x y a b

C = coeffs(x^2 + y^2 - (x + a*y)*(x + b*y),x)

C =

[a,b] = solve(C == 0,a,b)

a =

b =

Thus finding the two solutions. Pick either one to factor the original polynomial.

As I said, a hack. But perhaps another idea is to non-dimensionalize the problem, factoring out the highest order of one variable. Essentially, divide by y^2 below. So we would have

P = x^2 + y^2

P =

Q = P/y^2;

syms u

Q = simplify(subs(Q,x,u*y))

Q =

QF = factor(Q,'FactorMode','full')

QF =

simplify(y*subs(QF,u,x/y))

ans =

This one seems a little more general, though still arguably a hack. It can at least work for bivariate polynomials.

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yogeshwari patel 2023-12-2

Before putting up the question I already gone through the above mention answer. Why we cant direct use the comammnad why to divide the expression by y^2 .Convert it into one variable and then substitute again.

Is it so that it didnt work for two variable ?