Do Cubics always have one real root?












1












$begingroup$


I've seen a few conflicting pieces of information online.



So far, I know that with real coefficients there will always be one real root. But how about with complex coefficients?



At very least could you give me a counter example? a cubic with no real roots






polynomials complex-numbers roots real-numbers







share|cite|improve this question









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user7971589 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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    1












    $begingroup$


    I've seen a few conflicting pieces of information online.



    So far, I know that with real coefficients there will always be one real root. But how about with complex coefficients?



    At very least could you give me a counter example? a cubic with no real roots






    polynomials complex-numbers roots real-numbers







    share|cite|improve this question









    New contributor




    user7971589 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      1












      1








      1





      $begingroup$


      I've seen a few conflicting pieces of information online.



      So far, I know that with real coefficients there will always be one real root. But how about with complex coefficients?



      At very least could you give me a counter example? a cubic with no real roots






      polynomials complex-numbers roots real-numbers







      share|cite|improve this question









      New contributor




      user7971589 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      I've seen a few conflicting pieces of information online.



      So far, I know that with real coefficients there will always be one real root. But how about with complex coefficients?



      At very least could you give me a counter example? a cubic with no real roots






      polynomials complex-numbers roots real-numbers




      polynomials complex-numbers roots real-numbers






      share|cite|improve this question









      New contributor




      user7971589 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question









      New contributor




      user7971589 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|cite|improve this question




      share|cite|improve this question








      edited 3 hours ago









      Servaes

      27.8k34098




      27.8k34098






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      asked 3 hours ago









      user7971589user7971589

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          2 Answers
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          $begingroup$

          One of the best things you can remember is that over a field (like the reals or complex numbers) roots come from linear factors. Use this to build your own examples: $f(z) =(z-i)^3$. If you want three distinct complex roots, do something like $f(z) = (z-i)(z+i)(z-2i)$.






          share|cite|improve this answer









          $endgroup$





















            3












            $begingroup$

            As you already know, a cubic with real coefficients always has one real root, so there is no counterexample of a cubic with real coefficients with no real roots.



            A cubic with complex coefficients with no real roots is easy to find; take $x^3+i$.






            share|cite|improve this answer









            $endgroup$













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              2 Answers
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              4












              $begingroup$

              One of the best things you can remember is that over a field (like the reals or complex numbers) roots come from linear factors. Use this to build your own examples: $f(z) =(z-i)^3$. If you want three distinct complex roots, do something like $f(z) = (z-i)(z+i)(z-2i)$.






              share|cite|improve this answer









              $endgroup$


















                4












                $begingroup$

                One of the best things you can remember is that over a field (like the reals or complex numbers) roots come from linear factors. Use this to build your own examples: $f(z) =(z-i)^3$. If you want three distinct complex roots, do something like $f(z) = (z-i)(z+i)(z-2i)$.






                share|cite|improve this answer









                $endgroup$
















                  4












                  4








                  4





                  $begingroup$

                  One of the best things you can remember is that over a field (like the reals or complex numbers) roots come from linear factors. Use this to build your own examples: $f(z) =(z-i)^3$. If you want three distinct complex roots, do something like $f(z) = (z-i)(z+i)(z-2i)$.






                  share|cite|improve this answer









                  $endgroup$



                  One of the best things you can remember is that over a field (like the reals or complex numbers) roots come from linear factors. Use this to build your own examples: $f(z) =(z-i)^3$. If you want three distinct complex roots, do something like $f(z) = (z-i)(z+i)(z-2i)$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 3 hours ago









                  RandallRandall

                  10.3k11230




                  10.3k11230























                      3












                      $begingroup$

                      As you already know, a cubic with real coefficients always has one real root, so there is no counterexample of a cubic with real coefficients with no real roots.



                      A cubic with complex coefficients with no real roots is easy to find; take $x^3+i$.






                      share|cite|improve this answer









                      $endgroup$


















                        3












                        $begingroup$

                        As you already know, a cubic with real coefficients always has one real root, so there is no counterexample of a cubic with real coefficients with no real roots.



                        A cubic with complex coefficients with no real roots is easy to find; take $x^3+i$.






                        share|cite|improve this answer









                        $endgroup$
















                          3












                          3








                          3





                          $begingroup$

                          As you already know, a cubic with real coefficients always has one real root, so there is no counterexample of a cubic with real coefficients with no real roots.



                          A cubic with complex coefficients with no real roots is easy to find; take $x^3+i$.






                          share|cite|improve this answer









                          $endgroup$



                          As you already know, a cubic with real coefficients always has one real root, so there is no counterexample of a cubic with real coefficients with no real roots.



                          A cubic with complex coefficients with no real roots is easy to find; take $x^3+i$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 3 hours ago









                          ServaesServaes

                          27.8k34098




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