How to find the factors of the polynomial in one than one variable

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yogeshwari patel am 1 Dez. 2023
Beantwortet: Pavan Sahith am 13 Dez. 2023
syms x y
e1=x^2+4
factor(e1)
P = x^2 + y^2
QF = factor(P,x,'FactorMode','full')
factor(x^3 + 2, x, 'FactorMode', 'full')
I use the above code to find the complex factor but the answer are not correct
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yogeshwari patel am 1 Dez. 2023
for x^2+y^2 the factors are (x+iy) and (x-iy)
Bruno Luong am 1 Dez. 2023
Factor seems to be completly useless for such task, I ask the same question not long ago

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Antworten (1)

Pavan Sahith am 13 Dez. 2023
Hello Yogeswari,
As per my understanding, you want to factorize a polynomial in a complex field, and you are not able to get your expected output '(x+iy),(x-iy)' while using "factor(x^2+y^2,x,'FactorMode','full')".
When I tried with some equations like 'x^2+4',I could be able to get the linear factors.
syms x y
factor(x^2+ 4,x,"FactorMode","complex")
ans =
factor(x^3 + 2, x, 'FactorMode', 'full')
ans =
However, in the case of 'x^2+y^2', it appears that the answer remains irreducible. While searching the MathWorks documentation for "factor()", I found the following information:
"Factorization over complex numbers. A complex numeric factorization is a factorization into linear factors whose coefficients are floating-point numbers. Such factorization is only available if the coefficients of the input are convertible to floating-point numbers, that is, if the roots can be determined numerically. Symbolic inputs are treated as irreducible."
So,possibly 'y' might be getting considered as symbol and being treated as irreducable.
As a possible workaround, you might consider using the "solve" function:
syms x y;
e = x^2 + y^2;
roots = solve(e);
factors= prod(symvar(e,1) - roots)
factors =
Please find the related MathWorks documentation links below
Hope that Helps.
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