Factorize reciprocal polynomial 4th-order

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I try to factorize any polynomial like :



$x^4 + a.x^3 + b.x^2 + a.x + 1$ with $ a, b inBbbR$



into :



$(x^2 + c.x + d)(x^2 + e.x + f)$ with $ c, d, e, f inBbbR$



I also want $c(a, b)$, $d(a, b)$, $e(a, b)$, $f(a, b)$ to be continous, so I can be smooth at runtime when changing $(a, b)$.



The only way I succeeded for now is to compute all the roots and regroup them by conjugate pairs. But it's tricky because roots can be paired in many way, there are also cases where there are many solutions (when all roots are real), and I noticed roots may swap for specific values of $(a, b)$.



I would like to now if there is a simpler known method for this typical polynom.



Many thanks !










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  • Please use MathJax to type the equations. This way they will be much more understandable.
    – Matti P.
    14 hours ago










  • Thanks, edited.
    – user2443456
    14 hours ago










  • Well, just opening the parentheses in the second equation, we get $$ beginarrayrcc e + c &=& a \ f+d+ce &=& b \ cf + de &=& a \ df &=& 1 endarray $$
    – Matti P.
    14 hours ago















up vote
2
down vote

favorite
1












I try to factorize any polynomial like :



$x^4 + a.x^3 + b.x^2 + a.x + 1$ with $ a, b inBbbR$



into :



$(x^2 + c.x + d)(x^2 + e.x + f)$ with $ c, d, e, f inBbbR$



I also want $c(a, b)$, $d(a, b)$, $e(a, b)$, $f(a, b)$ to be continous, so I can be smooth at runtime when changing $(a, b)$.



The only way I succeeded for now is to compute all the roots and regroup them by conjugate pairs. But it's tricky because roots can be paired in many way, there are also cases where there are many solutions (when all roots are real), and I noticed roots may swap for specific values of $(a, b)$.



I would like to now if there is a simpler known method for this typical polynom.



Many thanks !










share|cite|improve this question























  • Please use MathJax to type the equations. This way they will be much more understandable.
    – Matti P.
    14 hours ago










  • Thanks, edited.
    – user2443456
    14 hours ago










  • Well, just opening the parentheses in the second equation, we get $$ beginarrayrcc e + c &=& a \ f+d+ce &=& b \ cf + de &=& a \ df &=& 1 endarray $$
    – Matti P.
    14 hours ago













up vote
2
down vote

favorite
1









up vote
2
down vote

favorite
1






1





I try to factorize any polynomial like :



$x^4 + a.x^3 + b.x^2 + a.x + 1$ with $ a, b inBbbR$



into :



$(x^2 + c.x + d)(x^2 + e.x + f)$ with $ c, d, e, f inBbbR$



I also want $c(a, b)$, $d(a, b)$, $e(a, b)$, $f(a, b)$ to be continous, so I can be smooth at runtime when changing $(a, b)$.



The only way I succeeded for now is to compute all the roots and regroup them by conjugate pairs. But it's tricky because roots can be paired in many way, there are also cases where there are many solutions (when all roots are real), and I noticed roots may swap for specific values of $(a, b)$.



I would like to now if there is a simpler known method for this typical polynom.



Many thanks !










share|cite|improve this question















I try to factorize any polynomial like :



$x^4 + a.x^3 + b.x^2 + a.x + 1$ with $ a, b inBbbR$



into :



$(x^2 + c.x + d)(x^2 + e.x + f)$ with $ c, d, e, f inBbbR$



I also want $c(a, b)$, $d(a, b)$, $e(a, b)$, $f(a, b)$ to be continous, so I can be smooth at runtime when changing $(a, b)$.



The only way I succeeded for now is to compute all the roots and regroup them by conjugate pairs. But it's tricky because roots can be paired in many way, there are also cases where there are many solutions (when all roots are real), and I noticed roots may swap for specific values of $(a, b)$.



I would like to now if there is a simpler known method for this typical polynom.



Many thanks !







polynomials






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edited 14 hours ago

























asked 14 hours ago









user2443456

374




374











  • Please use MathJax to type the equations. This way they will be much more understandable.
    – Matti P.
    14 hours ago










  • Thanks, edited.
    – user2443456
    14 hours ago










  • Well, just opening the parentheses in the second equation, we get $$ beginarrayrcc e + c &=& a \ f+d+ce &=& b \ cf + de &=& a \ df &=& 1 endarray $$
    – Matti P.
    14 hours ago

















  • Please use MathJax to type the equations. This way they will be much more understandable.
    – Matti P.
    14 hours ago










  • Thanks, edited.
    – user2443456
    14 hours ago










  • Well, just opening the parentheses in the second equation, we get $$ beginarrayrcc e + c &=& a \ f+d+ce &=& b \ cf + de &=& a \ df &=& 1 endarray $$
    – Matti P.
    14 hours ago
















Please use MathJax to type the equations. This way they will be much more understandable.
– Matti P.
14 hours ago




Please use MathJax to type the equations. This way they will be much more understandable.
– Matti P.
14 hours ago












Thanks, edited.
– user2443456
14 hours ago




Thanks, edited.
– user2443456
14 hours ago












Well, just opening the parentheses in the second equation, we get $$ beginarrayrcc e + c &=& a \ f+d+ce &=& b \ cf + de &=& a \ df &=& 1 endarray $$
– Matti P.
14 hours ago





Well, just opening the parentheses in the second equation, we get $$ beginarrayrcc e + c &=& a \ f+d+ce &=& b \ cf + de &=& a \ df &=& 1 endarray $$
– Matti P.
14 hours ago











2 Answers
2






active

oldest

votes

















up vote
4
down vote



accepted










As you pointed out yourself, the polynomial is palindromic (=equal to its own reciprocal). This means that $1/alpha$ is a zero whenever $alpha$ is.



In the case of four real roots this suggests using factors like $(x+alpha)(x+1/alpha)$. Such a factor has constant term $=1$, so you will have $d=f=1$.
This reduces your system of equations to $c+e=a$, $2+ce=b$. Meaning that $c$ and $e$ are the roots of the quadratic
$$
(T-c)(T-e)=T^2-(c+e)T+ce=T^2-aT+(b-2).
$$

The roots of this quadratic are real whenever we have the inequality
$$
a^2>4(b-2),
$$

and
$$
c,e=fracapmsqrta^2-4b+82.
$$




If this equality is not satisfied then you can proceed as follows.



  • Solve for $c$ and $e$ anyway (quadratic formula).

  • Find the zeros of $x^2+cx+1$ and $x^2+ex+1$ (quadratic formula again).

  • Match the complex zeros into conjugate pairs (one from each of the above pairs of solutions).

  • Build the factors.


Of course, when you have four real zeros you have many options with the factorization. The one I suggested is just easy to find, and the coefficients $c,d(,e,f)$ are easy to write down as functions of $a$ and $b$.






share|cite|improve this answer





























    up vote
    2
    down vote













    Write $$x^4+a x^3+b x^2+a x+1-left(x^2+c x+dright) left(x^2+e x+fright)=0$$ Expand and group terms to get
    $$(1-d f)+x (a-c f-d e)+x^2 (b-c e-d-f)+x^3 (a-c-e)=0$$ Set each coefficient equal to $0$ and you should get four nasty equations in $c,d,e,f$






    share|cite|improve this answer




















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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      4
      down vote



      accepted










      As you pointed out yourself, the polynomial is palindromic (=equal to its own reciprocal). This means that $1/alpha$ is a zero whenever $alpha$ is.



      In the case of four real roots this suggests using factors like $(x+alpha)(x+1/alpha)$. Such a factor has constant term $=1$, so you will have $d=f=1$.
      This reduces your system of equations to $c+e=a$, $2+ce=b$. Meaning that $c$ and $e$ are the roots of the quadratic
      $$
      (T-c)(T-e)=T^2-(c+e)T+ce=T^2-aT+(b-2).
      $$

      The roots of this quadratic are real whenever we have the inequality
      $$
      a^2>4(b-2),
      $$

      and
      $$
      c,e=fracapmsqrta^2-4b+82.
      $$




      If this equality is not satisfied then you can proceed as follows.



      • Solve for $c$ and $e$ anyway (quadratic formula).

      • Find the zeros of $x^2+cx+1$ and $x^2+ex+1$ (quadratic formula again).

      • Match the complex zeros into conjugate pairs (one from each of the above pairs of solutions).

      • Build the factors.


      Of course, when you have four real zeros you have many options with the factorization. The one I suggested is just easy to find, and the coefficients $c,d(,e,f)$ are easy to write down as functions of $a$ and $b$.






      share|cite|improve this answer


























        up vote
        4
        down vote



        accepted










        As you pointed out yourself, the polynomial is palindromic (=equal to its own reciprocal). This means that $1/alpha$ is a zero whenever $alpha$ is.



        In the case of four real roots this suggests using factors like $(x+alpha)(x+1/alpha)$. Such a factor has constant term $=1$, so you will have $d=f=1$.
        This reduces your system of equations to $c+e=a$, $2+ce=b$. Meaning that $c$ and $e$ are the roots of the quadratic
        $$
        (T-c)(T-e)=T^2-(c+e)T+ce=T^2-aT+(b-2).
        $$

        The roots of this quadratic are real whenever we have the inequality
        $$
        a^2>4(b-2),
        $$

        and
        $$
        c,e=fracapmsqrta^2-4b+82.
        $$




        If this equality is not satisfied then you can proceed as follows.



        • Solve for $c$ and $e$ anyway (quadratic formula).

        • Find the zeros of $x^2+cx+1$ and $x^2+ex+1$ (quadratic formula again).

        • Match the complex zeros into conjugate pairs (one from each of the above pairs of solutions).

        • Build the factors.


        Of course, when you have four real zeros you have many options with the factorization. The one I suggested is just easy to find, and the coefficients $c,d(,e,f)$ are easy to write down as functions of $a$ and $b$.






        share|cite|improve this answer
























          up vote
          4
          down vote



          accepted







          up vote
          4
          down vote



          accepted






          As you pointed out yourself, the polynomial is palindromic (=equal to its own reciprocal). This means that $1/alpha$ is a zero whenever $alpha$ is.



          In the case of four real roots this suggests using factors like $(x+alpha)(x+1/alpha)$. Such a factor has constant term $=1$, so you will have $d=f=1$.
          This reduces your system of equations to $c+e=a$, $2+ce=b$. Meaning that $c$ and $e$ are the roots of the quadratic
          $$
          (T-c)(T-e)=T^2-(c+e)T+ce=T^2-aT+(b-2).
          $$

          The roots of this quadratic are real whenever we have the inequality
          $$
          a^2>4(b-2),
          $$

          and
          $$
          c,e=fracapmsqrta^2-4b+82.
          $$




          If this equality is not satisfied then you can proceed as follows.



          • Solve for $c$ and $e$ anyway (quadratic formula).

          • Find the zeros of $x^2+cx+1$ and $x^2+ex+1$ (quadratic formula again).

          • Match the complex zeros into conjugate pairs (one from each of the above pairs of solutions).

          • Build the factors.


          Of course, when you have four real zeros you have many options with the factorization. The one I suggested is just easy to find, and the coefficients $c,d(,e,f)$ are easy to write down as functions of $a$ and $b$.






          share|cite|improve this answer














          As you pointed out yourself, the polynomial is palindromic (=equal to its own reciprocal). This means that $1/alpha$ is a zero whenever $alpha$ is.



          In the case of four real roots this suggests using factors like $(x+alpha)(x+1/alpha)$. Such a factor has constant term $=1$, so you will have $d=f=1$.
          This reduces your system of equations to $c+e=a$, $2+ce=b$. Meaning that $c$ and $e$ are the roots of the quadratic
          $$
          (T-c)(T-e)=T^2-(c+e)T+ce=T^2-aT+(b-2).
          $$

          The roots of this quadratic are real whenever we have the inequality
          $$
          a^2>4(b-2),
          $$

          and
          $$
          c,e=fracapmsqrta^2-4b+82.
          $$




          If this equality is not satisfied then you can proceed as follows.



          • Solve for $c$ and $e$ anyway (quadratic formula).

          • Find the zeros of $x^2+cx+1$ and $x^2+ex+1$ (quadratic formula again).

          • Match the complex zeros into conjugate pairs (one from each of the above pairs of solutions).

          • Build the factors.


          Of course, when you have four real zeros you have many options with the factorization. The one I suggested is just easy to find, and the coefficients $c,d(,e,f)$ are easy to write down as functions of $a$ and $b$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 13 hours ago

























          answered 13 hours ago









          Jyrki Lahtonen

          107k12166363




          107k12166363




















              up vote
              2
              down vote













              Write $$x^4+a x^3+b x^2+a x+1-left(x^2+c x+dright) left(x^2+e x+fright)=0$$ Expand and group terms to get
              $$(1-d f)+x (a-c f-d e)+x^2 (b-c e-d-f)+x^3 (a-c-e)=0$$ Set each coefficient equal to $0$ and you should get four nasty equations in $c,d,e,f$






              share|cite|improve this answer
























                up vote
                2
                down vote













                Write $$x^4+a x^3+b x^2+a x+1-left(x^2+c x+dright) left(x^2+e x+fright)=0$$ Expand and group terms to get
                $$(1-d f)+x (a-c f-d e)+x^2 (b-c e-d-f)+x^3 (a-c-e)=0$$ Set each coefficient equal to $0$ and you should get four nasty equations in $c,d,e,f$






                share|cite|improve this answer






















                  up vote
                  2
                  down vote










                  up vote
                  2
                  down vote









                  Write $$x^4+a x^3+b x^2+a x+1-left(x^2+c x+dright) left(x^2+e x+fright)=0$$ Expand and group terms to get
                  $$(1-d f)+x (a-c f-d e)+x^2 (b-c e-d-f)+x^3 (a-c-e)=0$$ Set each coefficient equal to $0$ and you should get four nasty equations in $c,d,e,f$






                  share|cite|improve this answer












                  Write $$x^4+a x^3+b x^2+a x+1-left(x^2+c x+dright) left(x^2+e x+fright)=0$$ Expand and group terms to get
                  $$(1-d f)+x (a-c f-d e)+x^2 (b-c e-d-f)+x^3 (a-c-e)=0$$ Set each coefficient equal to $0$ and you should get four nasty equations in $c,d,e,f$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 14 hours ago









                  Claude Leibovici

                  115k1155130




                  115k1155130



























                       

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