Decomposition of a polynomial
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My question : Given a polynomial
$$ sum_n=0^Na_nx^n $$
cannot be solved in general, but depending on the coefficients, can we know if it can be decomposed as a product of smaller order polynomials of degree 1 ,2 ,3 and 4 so in this case can be factorized and solved?
polynomials
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up vote
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down vote
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My question : Given a polynomial
$$ sum_n=0^Na_nx^n $$
cannot be solved in general, but depending on the coefficients, can we know if it can be decomposed as a product of smaller order polynomials of degree 1 ,2 ,3 and 4 so in this case can be factorized and solved?
polynomials
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
My question : Given a polynomial
$$ sum_n=0^Na_nx^n $$
cannot be solved in general, but depending on the coefficients, can we know if it can be decomposed as a product of smaller order polynomials of degree 1 ,2 ,3 and 4 so in this case can be factorized and solved?
polynomials
My question : Given a polynomial
$$ sum_n=0^Na_nx^n $$
cannot be solved in general, but depending on the coefficients, can we know if it can be decomposed as a product of smaller order polynomials of degree 1 ,2 ,3 and 4 so in this case can be factorized and solved?
polynomials
edited Aug 26 at 12:32
Bernard
111k635102
111k635102
asked Aug 26 at 11:30
Jose Garcia
4,00711235
4,00711235
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By the fundamental theorem of algebra we know that every non-zero $n^textth$ degree polynomial with constant complex coefficients has, counted with their algebraic multiplicity, exactly $n$ roots.
So by this means we can decompose, in the complex field, every polynomial into a product of polynomials with degree less than $n$. As an example take a $5^textth$ degree polynomial $P(x)$ with three roots $x_1, x_2, x_3$ such that $$texta.m.(x_1) = 1 ;; texta.m.(x_2) = 2;; texta.m.(x_3) = 2$$ were $texta.m.(x_n)$ is the algebraic multiplicity of $x_n$, then we can write $$P(x) = (x-x_1)(x-x_2)^2(x-x_3)^2$$
Depending on the coefficients you can say something about the roots of the polynomial by using Decartes' rule of signs or with the Routh-Hurwitz critetion. I don't think that, in a general case, you can say anything more than this, but I stand to be corrected. Surely we can say something more by knowing the form of the polynomial
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
By the fundamental theorem of algebra we know that every non-zero $n^textth$ degree polynomial with constant complex coefficients has, counted with their algebraic multiplicity, exactly $n$ roots.
So by this means we can decompose, in the complex field, every polynomial into a product of polynomials with degree less than $n$. As an example take a $5^textth$ degree polynomial $P(x)$ with three roots $x_1, x_2, x_3$ such that $$texta.m.(x_1) = 1 ;; texta.m.(x_2) = 2;; texta.m.(x_3) = 2$$ were $texta.m.(x_n)$ is the algebraic multiplicity of $x_n$, then we can write $$P(x) = (x-x_1)(x-x_2)^2(x-x_3)^2$$
Depending on the coefficients you can say something about the roots of the polynomial by using Decartes' rule of signs or with the Routh-Hurwitz critetion. I don't think that, in a general case, you can say anything more than this, but I stand to be corrected. Surely we can say something more by knowing the form of the polynomial
add a comment |Â
up vote
0
down vote
By the fundamental theorem of algebra we know that every non-zero $n^textth$ degree polynomial with constant complex coefficients has, counted with their algebraic multiplicity, exactly $n$ roots.
So by this means we can decompose, in the complex field, every polynomial into a product of polynomials with degree less than $n$. As an example take a $5^textth$ degree polynomial $P(x)$ with three roots $x_1, x_2, x_3$ such that $$texta.m.(x_1) = 1 ;; texta.m.(x_2) = 2;; texta.m.(x_3) = 2$$ were $texta.m.(x_n)$ is the algebraic multiplicity of $x_n$, then we can write $$P(x) = (x-x_1)(x-x_2)^2(x-x_3)^2$$
Depending on the coefficients you can say something about the roots of the polynomial by using Decartes' rule of signs or with the Routh-Hurwitz critetion. I don't think that, in a general case, you can say anything more than this, but I stand to be corrected. Surely we can say something more by knowing the form of the polynomial
add a comment |Â
up vote
0
down vote
up vote
0
down vote
By the fundamental theorem of algebra we know that every non-zero $n^textth$ degree polynomial with constant complex coefficients has, counted with their algebraic multiplicity, exactly $n$ roots.
So by this means we can decompose, in the complex field, every polynomial into a product of polynomials with degree less than $n$. As an example take a $5^textth$ degree polynomial $P(x)$ with three roots $x_1, x_2, x_3$ such that $$texta.m.(x_1) = 1 ;; texta.m.(x_2) = 2;; texta.m.(x_3) = 2$$ were $texta.m.(x_n)$ is the algebraic multiplicity of $x_n$, then we can write $$P(x) = (x-x_1)(x-x_2)^2(x-x_3)^2$$
Depending on the coefficients you can say something about the roots of the polynomial by using Decartes' rule of signs or with the Routh-Hurwitz critetion. I don't think that, in a general case, you can say anything more than this, but I stand to be corrected. Surely we can say something more by knowing the form of the polynomial
By the fundamental theorem of algebra we know that every non-zero $n^textth$ degree polynomial with constant complex coefficients has, counted with their algebraic multiplicity, exactly $n$ roots.
So by this means we can decompose, in the complex field, every polynomial into a product of polynomials with degree less than $n$. As an example take a $5^textth$ degree polynomial $P(x)$ with three roots $x_1, x_2, x_3$ such that $$texta.m.(x_1) = 1 ;; texta.m.(x_2) = 2;; texta.m.(x_3) = 2$$ were $texta.m.(x_n)$ is the algebraic multiplicity of $x_n$, then we can write $$P(x) = (x-x_1)(x-x_2)^2(x-x_3)^2$$
Depending on the coefficients you can say something about the roots of the polynomial by using Decartes' rule of signs or with the Routh-Hurwitz critetion. I don't think that, in a general case, you can say anything more than this, but I stand to be corrected. Surely we can say something more by knowing the form of the polynomial
edited Aug 26 at 11:47
answered Aug 26 at 11:41
Davide Morgante
2,513623
2,513623
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