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?







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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?







    share|cite|improve this question
























      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?







      share|cite|improve this question














      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?









      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Aug 26 at 12:32









      Bernard

      111k635102




      111k635102










      asked Aug 26 at 11:30









      Jose Garcia

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      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
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            1 Answer
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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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              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






              share|cite|improve this answer
























                up vote
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                up vote
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                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






                share|cite|improve this answer














                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







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Aug 26 at 11:47

























                answered Aug 26 at 11:41









                Davide Morgante

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