Real solutions of a monic polynomial $f(x) in mathbb R[x]$ only containing even degree terms












1












$begingroup$


Suppose $f(x) = x^{2k} + a_{2k-2} x^{2k-2} + dots + a_2 x^2 + a_0 in mathbb R[x]$ with $k in mathbb N$. Let us assume $k ge 2$. What can be said about the number of distinct real roots for $f(x)$?



By Fundamental Theorem of Algebra, it could be at most $2k$. But on the other hand, when I was playing with $x^4 + ax^2 + b$, it seems that if the roots are all real, it must be factored as $(x^2 - c^2)^2$ for some $c$ which would only give us two distinct real roots.






abstract-algebra polynomials roots







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    1












    $begingroup$


    Suppose $f(x) = x^{2k} + a_{2k-2} x^{2k-2} + dots + a_2 x^2 + a_0 in mathbb R[x]$ with $k in mathbb N$. Let us assume $k ge 2$. What can be said about the number of distinct real roots for $f(x)$?



    By Fundamental Theorem of Algebra, it could be at most $2k$. But on the other hand, when I was playing with $x^4 + ax^2 + b$, it seems that if the roots are all real, it must be factored as $(x^2 - c^2)^2$ for some $c$ which would only give us two distinct real roots.






    abstract-algebra polynomials roots







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Suppose $f(x) = x^{2k} + a_{2k-2} x^{2k-2} + dots + a_2 x^2 + a_0 in mathbb R[x]$ with $k in mathbb N$. Let us assume $k ge 2$. What can be said about the number of distinct real roots for $f(x)$?



      By Fundamental Theorem of Algebra, it could be at most $2k$. But on the other hand, when I was playing with $x^4 + ax^2 + b$, it seems that if the roots are all real, it must be factored as $(x^2 - c^2)^2$ for some $c$ which would only give us two distinct real roots.






      abstract-algebra polynomials roots







      share|cite|improve this question









      $endgroup$




      Suppose $f(x) = x^{2k} + a_{2k-2} x^{2k-2} + dots + a_2 x^2 + a_0 in mathbb R[x]$ with $k in mathbb N$. Let us assume $k ge 2$. What can be said about the number of distinct real roots for $f(x)$?



      By Fundamental Theorem of Algebra, it could be at most $2k$. But on the other hand, when I was playing with $x^4 + ax^2 + b$, it seems that if the roots are all real, it must be factored as $(x^2 - c^2)^2$ for some $c$ which would only give us two distinct real roots.






      abstract-algebra polynomials roots




      abstract-algebra polynomials roots






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      asked Jan 11 at 22:58









      user1101010user1101010

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          $begingroup$

          Well, writing $g(x)=x^k+a_{2k-2}x^{k-1}+dots+a_2x+a_0$, we have $f(x)=g(x^2)$. So, the roots of $f$ are just the square roots of the roots of $g$. This means that $f$ has two real roots for every positive root of $g$, and one more real root if $0$ is a root of $g$.



          In particular, $f$ could have any number of distinct real roots from $0$ to $2k$. Indeed, to get any even number $2m$ of real roots ($0leq mleq k$), just let $g$ be a polynomial with exactly $m$ positive roots and for which $0$ is not a root. To get an odd number $2m+1$ of real roots ($0leq m<k$), just let $g$ have $m$ positive roots and also have $0$ as a root.






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            2












            $begingroup$

            If $a_0ne 0$ then the number of real roots is even.

            If $a_0=0$, then the number of real roots is odd.



            Observe that $f(x)$ can be written as $g(x^2)$ for a polynomial $g$ (with the same coefficients $a_{2k}$).

            Every positive root $s$ of $g$ determines two real solutions of $f$: $pmsqrt s$.






            share|cite|improve this answer









            $endgroup$





















              2












              $begingroup$

              What can be said easily is that the real roots come in pairs, as obviously with $x_0$ also $-x_0$ will be a root, and if $0$ is a root, it has an even degree $ge 2$.



              Also, $x^4-5x^2+4=(x^2-1)(x^2-4)=0$ has roots $pm1, pm2$.



              Basically your equation has real roots $x_{2i-1}=-x_{2i}$ that correspond to the non-negative roots $y_i$ of $y^k+a_{2k-2}y^{k-1}+ldots+a_2y+a_0=0$ via $x^2_{2i-1}=x^2_{2i}=y_i$.






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                3 Answers
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                2












                $begingroup$

                Well, writing $g(x)=x^k+a_{2k-2}x^{k-1}+dots+a_2x+a_0$, we have $f(x)=g(x^2)$. So, the roots of $f$ are just the square roots of the roots of $g$. This means that $f$ has two real roots for every positive root of $g$, and one more real root if $0$ is a root of $g$.



                In particular, $f$ could have any number of distinct real roots from $0$ to $2k$. Indeed, to get any even number $2m$ of real roots ($0leq mleq k$), just let $g$ be a polynomial with exactly $m$ positive roots and for which $0$ is not a root. To get an odd number $2m+1$ of real roots ($0leq m<k$), just let $g$ have $m$ positive roots and also have $0$ as a root.






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  Well, writing $g(x)=x^k+a_{2k-2}x^{k-1}+dots+a_2x+a_0$, we have $f(x)=g(x^2)$. So, the roots of $f$ are just the square roots of the roots of $g$. This means that $f$ has two real roots for every positive root of $g$, and one more real root if $0$ is a root of $g$.



                  In particular, $f$ could have any number of distinct real roots from $0$ to $2k$. Indeed, to get any even number $2m$ of real roots ($0leq mleq k$), just let $g$ be a polynomial with exactly $m$ positive roots and for which $0$ is not a root. To get an odd number $2m+1$ of real roots ($0leq m<k$), just let $g$ have $m$ positive roots and also have $0$ as a root.






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    Well, writing $g(x)=x^k+a_{2k-2}x^{k-1}+dots+a_2x+a_0$, we have $f(x)=g(x^2)$. So, the roots of $f$ are just the square roots of the roots of $g$. This means that $f$ has two real roots for every positive root of $g$, and one more real root if $0$ is a root of $g$.



                    In particular, $f$ could have any number of distinct real roots from $0$ to $2k$. Indeed, to get any even number $2m$ of real roots ($0leq mleq k$), just let $g$ be a polynomial with exactly $m$ positive roots and for which $0$ is not a root. To get an odd number $2m+1$ of real roots ($0leq m<k$), just let $g$ have $m$ positive roots and also have $0$ as a root.






                    share|cite|improve this answer









                    $endgroup$



                    Well, writing $g(x)=x^k+a_{2k-2}x^{k-1}+dots+a_2x+a_0$, we have $f(x)=g(x^2)$. So, the roots of $f$ are just the square roots of the roots of $g$. This means that $f$ has two real roots for every positive root of $g$, and one more real root if $0$ is a root of $g$.



                    In particular, $f$ could have any number of distinct real roots from $0$ to $2k$. Indeed, to get any even number $2m$ of real roots ($0leq mleq k$), just let $g$ be a polynomial with exactly $m$ positive roots and for which $0$ is not a root. To get an odd number $2m+1$ of real roots ($0leq m<k$), just let $g$ have $m$ positive roots and also have $0$ as a root.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 11 at 23:08









                    Eric WofseyEric Wofsey

                    189k14216347




                    189k14216347























                        2












                        $begingroup$

                        If $a_0ne 0$ then the number of real roots is even.

                        If $a_0=0$, then the number of real roots is odd.



                        Observe that $f(x)$ can be written as $g(x^2)$ for a polynomial $g$ (with the same coefficients $a_{2k}$).

                        Every positive root $s$ of $g$ determines two real solutions of $f$: $pmsqrt s$.






                        share|cite|improve this answer









                        $endgroup$


















                          2












                          $begingroup$

                          If $a_0ne 0$ then the number of real roots is even.

                          If $a_0=0$, then the number of real roots is odd.



                          Observe that $f(x)$ can be written as $g(x^2)$ for a polynomial $g$ (with the same coefficients $a_{2k}$).

                          Every positive root $s$ of $g$ determines two real solutions of $f$: $pmsqrt s$.






                          share|cite|improve this answer









                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            If $a_0ne 0$ then the number of real roots is even.

                            If $a_0=0$, then the number of real roots is odd.



                            Observe that $f(x)$ can be written as $g(x^2)$ for a polynomial $g$ (with the same coefficients $a_{2k}$).

                            Every positive root $s$ of $g$ determines two real solutions of $f$: $pmsqrt s$.






                            share|cite|improve this answer









                            $endgroup$



                            If $a_0ne 0$ then the number of real roots is even.

                            If $a_0=0$, then the number of real roots is odd.



                            Observe that $f(x)$ can be written as $g(x^2)$ for a polynomial $g$ (with the same coefficients $a_{2k}$).

                            Every positive root $s$ of $g$ determines two real solutions of $f$: $pmsqrt s$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 11 at 23:07









                            BerciBerci

                            61.3k23674




                            61.3k23674























                                2












                                $begingroup$

                                What can be said easily is that the real roots come in pairs, as obviously with $x_0$ also $-x_0$ will be a root, and if $0$ is a root, it has an even degree $ge 2$.



                                Also, $x^4-5x^2+4=(x^2-1)(x^2-4)=0$ has roots $pm1, pm2$.



                                Basically your equation has real roots $x_{2i-1}=-x_{2i}$ that correspond to the non-negative roots $y_i$ of $y^k+a_{2k-2}y^{k-1}+ldots+a_2y+a_0=0$ via $x^2_{2i-1}=x^2_{2i}=y_i$.






                                share|cite|improve this answer









                                $endgroup$


















                                  2












                                  $begingroup$

                                  What can be said easily is that the real roots come in pairs, as obviously with $x_0$ also $-x_0$ will be a root, and if $0$ is a root, it has an even degree $ge 2$.



                                  Also, $x^4-5x^2+4=(x^2-1)(x^2-4)=0$ has roots $pm1, pm2$.



                                  Basically your equation has real roots $x_{2i-1}=-x_{2i}$ that correspond to the non-negative roots $y_i$ of $y^k+a_{2k-2}y^{k-1}+ldots+a_2y+a_0=0$ via $x^2_{2i-1}=x^2_{2i}=y_i$.






                                  share|cite|improve this answer









                                  $endgroup$
















                                    2












                                    2








                                    2





                                    $begingroup$

                                    What can be said easily is that the real roots come in pairs, as obviously with $x_0$ also $-x_0$ will be a root, and if $0$ is a root, it has an even degree $ge 2$.



                                    Also, $x^4-5x^2+4=(x^2-1)(x^2-4)=0$ has roots $pm1, pm2$.



                                    Basically your equation has real roots $x_{2i-1}=-x_{2i}$ that correspond to the non-negative roots $y_i$ of $y^k+a_{2k-2}y^{k-1}+ldots+a_2y+a_0=0$ via $x^2_{2i-1}=x^2_{2i}=y_i$.






                                    share|cite|improve this answer









                                    $endgroup$



                                    What can be said easily is that the real roots come in pairs, as obviously with $x_0$ also $-x_0$ will be a root, and if $0$ is a root, it has an even degree $ge 2$.



                                    Also, $x^4-5x^2+4=(x^2-1)(x^2-4)=0$ has roots $pm1, pm2$.



                                    Basically your equation has real roots $x_{2i-1}=-x_{2i}$ that correspond to the non-negative roots $y_i$ of $y^k+a_{2k-2}y^{k-1}+ldots+a_2y+a_0=0$ via $x^2_{2i-1}=x^2_{2i}=y_i$.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Jan 11 at 23:10









                                    IngixIngix

                                    4,832159




                                    4,832159






























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