Can some cubic polynomial have two real roots?












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In $p(x) = x^3-x^2$, both $0$ and $1$ are possible roots of the polynomial; both are real. I had read that a cubic polynomial has either all real roots or just one real root. It can't have two. What is the problem in this case?






polynomials complex-numbers roots cubic-equations







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$endgroup$








  • 1




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    I think the statement is that if a cubic polynomial has three distinct roots, then either all 3 are real or only one of them is real and the other two are complex conjugates of each other.
    $endgroup$
    – harshit54
    Jan 2 at 12:45










  • $begingroup$
    The title is not good. A trivial answer would be: Yes, take $f(x)=x^3$. It has $3$ real roots, $0$, $0$ and again $0$.
    $endgroup$
    – Dietrich Burde
    Jan 2 at 13:46


















0












$begingroup$


In $p(x) = x^3-x^2$, both $0$ and $1$ are possible roots of the polynomial; both are real. I had read that a cubic polynomial has either all real roots or just one real root. It can't have two. What is the problem in this case?






polynomials complex-numbers roots cubic-equations







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I think the statement is that if a cubic polynomial has three distinct roots, then either all 3 are real or only one of them is real and the other two are complex conjugates of each other.
    $endgroup$
    – harshit54
    Jan 2 at 12:45










  • $begingroup$
    The title is not good. A trivial answer would be: Yes, take $f(x)=x^3$. It has $3$ real roots, $0$, $0$ and again $0$.
    $endgroup$
    – Dietrich Burde
    Jan 2 at 13:46
















0












0








0





$begingroup$


In $p(x) = x^3-x^2$, both $0$ and $1$ are possible roots of the polynomial; both are real. I had read that a cubic polynomial has either all real roots or just one real root. It can't have two. What is the problem in this case?






polynomials complex-numbers roots cubic-equations







share|cite|improve this question











$endgroup$




In $p(x) = x^3-x^2$, both $0$ and $1$ are possible roots of the polynomial; both are real. I had read that a cubic polynomial has either all real roots or just one real root. It can't have two. What is the problem in this case?






polynomials complex-numbers roots cubic-equations




polynomials complex-numbers roots cubic-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 14:01









Shailesh

3,95392134




3,95392134










asked Jan 2 at 12:40









RaenRaen

354




354








  • 1




    $begingroup$
    I think the statement is that if a cubic polynomial has three distinct roots, then either all 3 are real or only one of them is real and the other two are complex conjugates of each other.
    $endgroup$
    – harshit54
    Jan 2 at 12:45










  • $begingroup$
    The title is not good. A trivial answer would be: Yes, take $f(x)=x^3$. It has $3$ real roots, $0$, $0$ and again $0$.
    $endgroup$
    – Dietrich Burde
    Jan 2 at 13:46
















  • 1




    $begingroup$
    I think the statement is that if a cubic polynomial has three distinct roots, then either all 3 are real or only one of them is real and the other two are complex conjugates of each other.
    $endgroup$
    – harshit54
    Jan 2 at 12:45










  • $begingroup$
    The title is not good. A trivial answer would be: Yes, take $f(x)=x^3$. It has $3$ real roots, $0$, $0$ and again $0$.
    $endgroup$
    – Dietrich Burde
    Jan 2 at 13:46










1




1




$begingroup$
I think the statement is that if a cubic polynomial has three distinct roots, then either all 3 are real or only one of them is real and the other two are complex conjugates of each other.
$endgroup$
– harshit54
Jan 2 at 12:45




$begingroup$
I think the statement is that if a cubic polynomial has three distinct roots, then either all 3 are real or only one of them is real and the other two are complex conjugates of each other.
$endgroup$
– harshit54
Jan 2 at 12:45












$begingroup$
The title is not good. A trivial answer would be: Yes, take $f(x)=x^3$. It has $3$ real roots, $0$, $0$ and again $0$.
$endgroup$
– Dietrich Burde
Jan 2 at 13:46






$begingroup$
The title is not good. A trivial answer would be: Yes, take $f(x)=x^3$. It has $3$ real roots, $0$, $0$ and again $0$.
$endgroup$
– Dietrich Burde
Jan 2 at 13:46












3 Answers
3






active

oldest

votes


















5












$begingroup$

In your case, $0$ is a double root: you should count it as two roots. In other words, the following statement holds:




If the roots are counted with their multiplicities, then every cubic polynomial in one variable with real coefficients either has exactly one real root or it has three real roots.







share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Note: if you have complex coefficients instead, the statement isn’t true.



    As an example, consider $f(z) = (z-1)(z+1)(z-i)$. $f$ has two real roots, but one complex root.






    share|cite|improve this answer









    $endgroup$





















      -1












      $begingroup$

      The point is that complex conjugation ${Bbb C}rightarrow {Bbb C}:zmapsto bar z$, where $bar z = a-ib$ if $z=a+ib$, is a (ring) automorphism. It follows that if you have a polynomial (with real coefficients in your case) $f(x)$ with $f(z)=0$ for some $zin{Bbb C}$, then $f(bar z) = overline {f(z)} = bar 0 = 0$. Thus complex roots always occur in pairs: $(z,bar z)$.



      So actually a cubic polynomial with real coefficients can only have 1 or 3 real roots, but not 2. If if would have 2 real roots and 1 complex root $z$, then $bar z$ would also be a root and so (as argumented above) $z=bar z$ would be real.






      share|cite|improve this answer











      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        5












        $begingroup$

        In your case, $0$ is a double root: you should count it as two roots. In other words, the following statement holds:




        If the roots are counted with their multiplicities, then every cubic polynomial in one variable with real coefficients either has exactly one real root or it has three real roots.







        share|cite|improve this answer









        $endgroup$


















          5












          $begingroup$

          In your case, $0$ is a double root: you should count it as two roots. In other words, the following statement holds:




          If the roots are counted with their multiplicities, then every cubic polynomial in one variable with real coefficients either has exactly one real root or it has three real roots.







          share|cite|improve this answer









          $endgroup$
















            5












            5








            5





            $begingroup$

            In your case, $0$ is a double root: you should count it as two roots. In other words, the following statement holds:




            If the roots are counted with their multiplicities, then every cubic polynomial in one variable with real coefficients either has exactly one real root or it has three real roots.







            share|cite|improve this answer









            $endgroup$



            In your case, $0$ is a double root: you should count it as two roots. In other words, the following statement holds:




            If the roots are counted with their multiplicities, then every cubic polynomial in one variable with real coefficients either has exactly one real root or it has three real roots.








            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 2 at 12:43









            José Carlos SantosJosé Carlos Santos

            164k22131235




            164k22131235























                0












                $begingroup$

                Note: if you have complex coefficients instead, the statement isn’t true.



                As an example, consider $f(z) = (z-1)(z+1)(z-i)$. $f$ has two real roots, but one complex root.






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  Note: if you have complex coefficients instead, the statement isn’t true.



                  As an example, consider $f(z) = (z-1)(z+1)(z-i)$. $f$ has two real roots, but one complex root.






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    Note: if you have complex coefficients instead, the statement isn’t true.



                    As an example, consider $f(z) = (z-1)(z+1)(z-i)$. $f$ has two real roots, but one complex root.






                    share|cite|improve this answer









                    $endgroup$



                    Note: if you have complex coefficients instead, the statement isn’t true.



                    As an example, consider $f(z) = (z-1)(z+1)(z-i)$. $f$ has two real roots, but one complex root.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 2 at 15:37









                    user458276user458276

                    674211




                    674211























                        -1












                        $begingroup$

                        The point is that complex conjugation ${Bbb C}rightarrow {Bbb C}:zmapsto bar z$, where $bar z = a-ib$ if $z=a+ib$, is a (ring) automorphism. It follows that if you have a polynomial (with real coefficients in your case) $f(x)$ with $f(z)=0$ for some $zin{Bbb C}$, then $f(bar z) = overline {f(z)} = bar 0 = 0$. Thus complex roots always occur in pairs: $(z,bar z)$.



                        So actually a cubic polynomial with real coefficients can only have 1 or 3 real roots, but not 2. If if would have 2 real roots and 1 complex root $z$, then $bar z$ would also be a root and so (as argumented above) $z=bar z$ would be real.






                        share|cite|improve this answer











                        $endgroup$


















                          -1












                          $begingroup$

                          The point is that complex conjugation ${Bbb C}rightarrow {Bbb C}:zmapsto bar z$, where $bar z = a-ib$ if $z=a+ib$, is a (ring) automorphism. It follows that if you have a polynomial (with real coefficients in your case) $f(x)$ with $f(z)=0$ for some $zin{Bbb C}$, then $f(bar z) = overline {f(z)} = bar 0 = 0$. Thus complex roots always occur in pairs: $(z,bar z)$.



                          So actually a cubic polynomial with real coefficients can only have 1 or 3 real roots, but not 2. If if would have 2 real roots and 1 complex root $z$, then $bar z$ would also be a root and so (as argumented above) $z=bar z$ would be real.






                          share|cite|improve this answer











                          $endgroup$
















                            -1












                            -1








                            -1





                            $begingroup$

                            The point is that complex conjugation ${Bbb C}rightarrow {Bbb C}:zmapsto bar z$, where $bar z = a-ib$ if $z=a+ib$, is a (ring) automorphism. It follows that if you have a polynomial (with real coefficients in your case) $f(x)$ with $f(z)=0$ for some $zin{Bbb C}$, then $f(bar z) = overline {f(z)} = bar 0 = 0$. Thus complex roots always occur in pairs: $(z,bar z)$.



                            So actually a cubic polynomial with real coefficients can only have 1 or 3 real roots, but not 2. If if would have 2 real roots and 1 complex root $z$, then $bar z$ would also be a root and so (as argumented above) $z=bar z$ would be real.






                            share|cite|improve this answer











                            $endgroup$



                            The point is that complex conjugation ${Bbb C}rightarrow {Bbb C}:zmapsto bar z$, where $bar z = a-ib$ if $z=a+ib$, is a (ring) automorphism. It follows that if you have a polynomial (with real coefficients in your case) $f(x)$ with $f(z)=0$ for some $zin{Bbb C}$, then $f(bar z) = overline {f(z)} = bar 0 = 0$. Thus complex roots always occur in pairs: $(z,bar z)$.



                            So actually a cubic polynomial with real coefficients can only have 1 or 3 real roots, but not 2. If if would have 2 real roots and 1 complex root $z$, then $bar z$ would also be a root and so (as argumented above) $z=bar z$ would be real.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Jan 3 at 8:48

























                            answered Jan 2 at 12:50









                            WuestenfuxWuestenfux

                            4,7941513




                            4,7941513






























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