Values of $a$ for which $p(x)$ has a complex roots?












2














Let $p(x) = x^4 + ax^3 + bx^2 + ax + 1$. Suppose $x = 1$ is a root of $p(x)$, then find the range of values of $a$ such that $p(x)$ has complex (non-real) roots?



My approach: Using the fact that $1$ is a root, I am able to deduce $b = -2(a+1)$. Then I wrote $$p(x) = (x-1)(x-z)(x-bar{z})(x-k)$$ where $z = c + id, dneq 0$ and $k$ is some other real roots. Now I plan to expand and compare the coefficients but have been unable to do so in a fruitful manner (there just too many variables to get to a proper estimate of range of $a$ )






polynomials







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    2














    Let $p(x) = x^4 + ax^3 + bx^2 + ax + 1$. Suppose $x = 1$ is a root of $p(x)$, then find the range of values of $a$ such that $p(x)$ has complex (non-real) roots?



    My approach: Using the fact that $1$ is a root, I am able to deduce $b = -2(a+1)$. Then I wrote $$p(x) = (x-1)(x-z)(x-bar{z})(x-k)$$ where $z = c + id, dneq 0$ and $k$ is some other real roots. Now I plan to expand and compare the coefficients but have been unable to do so in a fruitful manner (there just too many variables to get to a proper estimate of range of $a$ )






    polynomials







    share|cite|improve this question

























      2












      2








      2


      1





      Let $p(x) = x^4 + ax^3 + bx^2 + ax + 1$. Suppose $x = 1$ is a root of $p(x)$, then find the range of values of $a$ such that $p(x)$ has complex (non-real) roots?



      My approach: Using the fact that $1$ is a root, I am able to deduce $b = -2(a+1)$. Then I wrote $$p(x) = (x-1)(x-z)(x-bar{z})(x-k)$$ where $z = c + id, dneq 0$ and $k$ is some other real roots. Now I plan to expand and compare the coefficients but have been unable to do so in a fruitful manner (there just too many variables to get to a proper estimate of range of $a$ )






      polynomials







      share|cite|improve this question













      Let $p(x) = x^4 + ax^3 + bx^2 + ax + 1$. Suppose $x = 1$ is a root of $p(x)$, then find the range of values of $a$ such that $p(x)$ has complex (non-real) roots?



      My approach: Using the fact that $1$ is a root, I am able to deduce $b = -2(a+1)$. Then I wrote $$p(x) = (x-1)(x-z)(x-bar{z})(x-k)$$ where $z = c + id, dneq 0$ and $k$ is some other real roots. Now I plan to expand and compare the coefficients but have been unable to do so in a fruitful manner (there just too many variables to get to a proper estimate of range of $a$ )






      polynomials




      polynomials






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      share|cite|improve this question











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      asked Nov 28 '18 at 1:37









      henceproved

      1358




      1358






















          4 Answers
          4






          active

          oldest

          votes


















          4














          HINT:



          $$p(x)=(x-1)(x^3+(a+1)x^2-(a+1)x-1)$$



          Rest follows very nicely.






          share|cite|improve this answer





















          • If you don't see how, theres a solution to $$x^3+(a+1)x^2-(a+1)x-1=0$$ that you should be able to spot immediately.
            – Rhys Hughes
            Nov 28 '18 at 2:03










          • $1$ is a mutiple root and after that I have to show that determinant is $leq 0$ for the remaining quadritic term.
            – henceproved
            Nov 28 '18 at 2:09










          • Absolutely, except the inequality is strict $(<0$ rather than $le 0)$
            – Rhys Hughes
            Nov 28 '18 at 5:09



















          1














          When a polynomial is "palindromic", reading the same way forwards and backwards, it must also have palindromic factors. Thus



          $x^4+ax^3+bx^2+ax+1=(x^2+cx+1)(x^2+dx+1)$



          If $x=1$ is to be a root then it must give $c=-2$ or $d=-2$ in one of the factors, as no other choice has the root $x=1$. Taking $c$ as $-2$ we then get the complex roots if $-2<d<2$, corresponding to a negative discriminate for the $x^2+dx+1$ factor.



          Expanding the product $(x^2+cx+1)(x^2+dx+1)$ and equating this to $x^4+ax^3+bx^2+ax+1$ then implies $a=c+d=d-2$ so the solution to the problem is $-4<a<0$.






          share|cite|improve this answer





























            0














            Here's a different approach. (I can't say for sure whether you can or cannot do it by comparing coefficients though!)



            For a degree 2 polynomial, we know that it has complex roots (assuming the coefficients are real) iff the discriminant is negative. There's a similar generalisation for high degree polynomial.



            In particular, a degree 4 polynomial
            $$ax^4+bx^3+cx^2+dx+e$$
            has discriminant
            $$256a^3e^3-192a^2bde^2-128a^2c^2e^2+144a^2d^2e...$$



            (see https://en.wikipedia.org/wiki/Discriminant#Degree_4 for the full formula). And we know that since $x=1$ is a root there can only be a pair of complex roots so we require the discriminant to be negative.






            share|cite|improve this answer





















            • I doubt this approach is feasible as I will be solving sixth degree $p(a)< 0$.
              – henceproved
              Nov 28 '18 at 1:56












            • Actually your polynomial always has a double root so the discriminant is always 0 for all $a$ (from which you can work from there). But anyway the above answer is definitely better.
              – Rcwt
              Nov 28 '18 at 2:04





















            0














            Considering that $$p(x) = x^4 + ax^3 + bx^2 + ax + 1$$ If $x=1$ is a root, then
            $$p(1)=2 a+b+2=0 implies b=-2-2a$$ and
            $$p(x)=(x-1),(x^3+(a+1) x^2-(a+1) x-1)$$ Clearly, $x=1$ is also a root for the second term making
            $$p(x)=(x-1)^2,(x^2+(a+2) x+1)$$ So, there are complex roots if the discriminant $Delta$ of the remaining quadratic equation is negative.






            share|cite|improve this answer





















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              4 Answers
              4






              active

              oldest

              votes








              4 Answers
              4






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              4














              HINT:



              $$p(x)=(x-1)(x^3+(a+1)x^2-(a+1)x-1)$$



              Rest follows very nicely.






              share|cite|improve this answer





















              • If you don't see how, theres a solution to $$x^3+(a+1)x^2-(a+1)x-1=0$$ that you should be able to spot immediately.
                – Rhys Hughes
                Nov 28 '18 at 2:03










              • $1$ is a mutiple root and after that I have to show that determinant is $leq 0$ for the remaining quadritic term.
                – henceproved
                Nov 28 '18 at 2:09










              • Absolutely, except the inequality is strict $(<0$ rather than $le 0)$
                – Rhys Hughes
                Nov 28 '18 at 5:09
















              4














              HINT:



              $$p(x)=(x-1)(x^3+(a+1)x^2-(a+1)x-1)$$



              Rest follows very nicely.






              share|cite|improve this answer





















              • If you don't see how, theres a solution to $$x^3+(a+1)x^2-(a+1)x-1=0$$ that you should be able to spot immediately.
                – Rhys Hughes
                Nov 28 '18 at 2:03










              • $1$ is a mutiple root and after that I have to show that determinant is $leq 0$ for the remaining quadritic term.
                – henceproved
                Nov 28 '18 at 2:09










              • Absolutely, except the inequality is strict $(<0$ rather than $le 0)$
                – Rhys Hughes
                Nov 28 '18 at 5:09














              4












              4








              4






              HINT:



              $$p(x)=(x-1)(x^3+(a+1)x^2-(a+1)x-1)$$



              Rest follows very nicely.






              share|cite|improve this answer












              HINT:



              $$p(x)=(x-1)(x^3+(a+1)x^2-(a+1)x-1)$$



              Rest follows very nicely.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 28 '18 at 1:58









              Rhys Hughes

              4,8041327




              4,8041327












              • If you don't see how, theres a solution to $$x^3+(a+1)x^2-(a+1)x-1=0$$ that you should be able to spot immediately.
                – Rhys Hughes
                Nov 28 '18 at 2:03










              • $1$ is a mutiple root and after that I have to show that determinant is $leq 0$ for the remaining quadritic term.
                – henceproved
                Nov 28 '18 at 2:09










              • Absolutely, except the inequality is strict $(<0$ rather than $le 0)$
                – Rhys Hughes
                Nov 28 '18 at 5:09


















              • If you don't see how, theres a solution to $$x^3+(a+1)x^2-(a+1)x-1=0$$ that you should be able to spot immediately.
                – Rhys Hughes
                Nov 28 '18 at 2:03










              • $1$ is a mutiple root and after that I have to show that determinant is $leq 0$ for the remaining quadritic term.
                – henceproved
                Nov 28 '18 at 2:09










              • Absolutely, except the inequality is strict $(<0$ rather than $le 0)$
                – Rhys Hughes
                Nov 28 '18 at 5:09
















              If you don't see how, theres a solution to $$x^3+(a+1)x^2-(a+1)x-1=0$$ that you should be able to spot immediately.
              – Rhys Hughes
              Nov 28 '18 at 2:03




              If you don't see how, theres a solution to $$x^3+(a+1)x^2-(a+1)x-1=0$$ that you should be able to spot immediately.
              – Rhys Hughes
              Nov 28 '18 at 2:03












              $1$ is a mutiple root and after that I have to show that determinant is $leq 0$ for the remaining quadritic term.
              – henceproved
              Nov 28 '18 at 2:09




              $1$ is a mutiple root and after that I have to show that determinant is $leq 0$ for the remaining quadritic term.
              – henceproved
              Nov 28 '18 at 2:09












              Absolutely, except the inequality is strict $(<0$ rather than $le 0)$
              – Rhys Hughes
              Nov 28 '18 at 5:09




              Absolutely, except the inequality is strict $(<0$ rather than $le 0)$
              – Rhys Hughes
              Nov 28 '18 at 5:09











              1














              When a polynomial is "palindromic", reading the same way forwards and backwards, it must also have palindromic factors. Thus



              $x^4+ax^3+bx^2+ax+1=(x^2+cx+1)(x^2+dx+1)$



              If $x=1$ is to be a root then it must give $c=-2$ or $d=-2$ in one of the factors, as no other choice has the root $x=1$. Taking $c$ as $-2$ we then get the complex roots if $-2<d<2$, corresponding to a negative discriminate for the $x^2+dx+1$ factor.



              Expanding the product $(x^2+cx+1)(x^2+dx+1)$ and equating this to $x^4+ax^3+bx^2+ax+1$ then implies $a=c+d=d-2$ so the solution to the problem is $-4<a<0$.






              share|cite|improve this answer


























                1














                When a polynomial is "palindromic", reading the same way forwards and backwards, it must also have palindromic factors. Thus



                $x^4+ax^3+bx^2+ax+1=(x^2+cx+1)(x^2+dx+1)$



                If $x=1$ is to be a root then it must give $c=-2$ or $d=-2$ in one of the factors, as no other choice has the root $x=1$. Taking $c$ as $-2$ we then get the complex roots if $-2<d<2$, corresponding to a negative discriminate for the $x^2+dx+1$ factor.



                Expanding the product $(x^2+cx+1)(x^2+dx+1)$ and equating this to $x^4+ax^3+bx^2+ax+1$ then implies $a=c+d=d-2$ so the solution to the problem is $-4<a<0$.






                share|cite|improve this answer
























                  1












                  1








                  1






                  When a polynomial is "palindromic", reading the same way forwards and backwards, it must also have palindromic factors. Thus



                  $x^4+ax^3+bx^2+ax+1=(x^2+cx+1)(x^2+dx+1)$



                  If $x=1$ is to be a root then it must give $c=-2$ or $d=-2$ in one of the factors, as no other choice has the root $x=1$. Taking $c$ as $-2$ we then get the complex roots if $-2<d<2$, corresponding to a negative discriminate for the $x^2+dx+1$ factor.



                  Expanding the product $(x^2+cx+1)(x^2+dx+1)$ and equating this to $x^4+ax^3+bx^2+ax+1$ then implies $a=c+d=d-2$ so the solution to the problem is $-4<a<0$.






                  share|cite|improve this answer












                  When a polynomial is "palindromic", reading the same way forwards and backwards, it must also have palindromic factors. Thus



                  $x^4+ax^3+bx^2+ax+1=(x^2+cx+1)(x^2+dx+1)$



                  If $x=1$ is to be a root then it must give $c=-2$ or $d=-2$ in one of the factors, as no other choice has the root $x=1$. Taking $c$ as $-2$ we then get the complex roots if $-2<d<2$, corresponding to a negative discriminate for the $x^2+dx+1$ factor.



                  Expanding the product $(x^2+cx+1)(x^2+dx+1)$ and equating this to $x^4+ax^3+bx^2+ax+1$ then implies $a=c+d=d-2$ so the solution to the problem is $-4<a<0$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 28 '18 at 2:04









                  Oscar Lanzi

                  12.1k12036




                  12.1k12036























                      0














                      Here's a different approach. (I can't say for sure whether you can or cannot do it by comparing coefficients though!)



                      For a degree 2 polynomial, we know that it has complex roots (assuming the coefficients are real) iff the discriminant is negative. There's a similar generalisation for high degree polynomial.



                      In particular, a degree 4 polynomial
                      $$ax^4+bx^3+cx^2+dx+e$$
                      has discriminant
                      $$256a^3e^3-192a^2bde^2-128a^2c^2e^2+144a^2d^2e...$$



                      (see https://en.wikipedia.org/wiki/Discriminant#Degree_4 for the full formula). And we know that since $x=1$ is a root there can only be a pair of complex roots so we require the discriminant to be negative.






                      share|cite|improve this answer





















                      • I doubt this approach is feasible as I will be solving sixth degree $p(a)< 0$.
                        – henceproved
                        Nov 28 '18 at 1:56












                      • Actually your polynomial always has a double root so the discriminant is always 0 for all $a$ (from which you can work from there). But anyway the above answer is definitely better.
                        – Rcwt
                        Nov 28 '18 at 2:04


















                      0














                      Here's a different approach. (I can't say for sure whether you can or cannot do it by comparing coefficients though!)



                      For a degree 2 polynomial, we know that it has complex roots (assuming the coefficients are real) iff the discriminant is negative. There's a similar generalisation for high degree polynomial.



                      In particular, a degree 4 polynomial
                      $$ax^4+bx^3+cx^2+dx+e$$
                      has discriminant
                      $$256a^3e^3-192a^2bde^2-128a^2c^2e^2+144a^2d^2e...$$



                      (see https://en.wikipedia.org/wiki/Discriminant#Degree_4 for the full formula). And we know that since $x=1$ is a root there can only be a pair of complex roots so we require the discriminant to be negative.






                      share|cite|improve this answer





















                      • I doubt this approach is feasible as I will be solving sixth degree $p(a)< 0$.
                        – henceproved
                        Nov 28 '18 at 1:56












                      • Actually your polynomial always has a double root so the discriminant is always 0 for all $a$ (from which you can work from there). But anyway the above answer is definitely better.
                        – Rcwt
                        Nov 28 '18 at 2:04
















                      0












                      0








                      0






                      Here's a different approach. (I can't say for sure whether you can or cannot do it by comparing coefficients though!)



                      For a degree 2 polynomial, we know that it has complex roots (assuming the coefficients are real) iff the discriminant is negative. There's a similar generalisation for high degree polynomial.



                      In particular, a degree 4 polynomial
                      $$ax^4+bx^3+cx^2+dx+e$$
                      has discriminant
                      $$256a^3e^3-192a^2bde^2-128a^2c^2e^2+144a^2d^2e...$$



                      (see https://en.wikipedia.org/wiki/Discriminant#Degree_4 for the full formula). And we know that since $x=1$ is a root there can only be a pair of complex roots so we require the discriminant to be negative.






                      share|cite|improve this answer












                      Here's a different approach. (I can't say for sure whether you can or cannot do it by comparing coefficients though!)



                      For a degree 2 polynomial, we know that it has complex roots (assuming the coefficients are real) iff the discriminant is negative. There's a similar generalisation for high degree polynomial.



                      In particular, a degree 4 polynomial
                      $$ax^4+bx^3+cx^2+dx+e$$
                      has discriminant
                      $$256a^3e^3-192a^2bde^2-128a^2c^2e^2+144a^2d^2e...$$



                      (see https://en.wikipedia.org/wiki/Discriminant#Degree_4 for the full formula). And we know that since $x=1$ is a root there can only be a pair of complex roots so we require the discriminant to be negative.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Nov 28 '18 at 1:50









                      Rcwt

                      3871210




                      3871210












                      • I doubt this approach is feasible as I will be solving sixth degree $p(a)< 0$.
                        – henceproved
                        Nov 28 '18 at 1:56












                      • Actually your polynomial always has a double root so the discriminant is always 0 for all $a$ (from which you can work from there). But anyway the above answer is definitely better.
                        – Rcwt
                        Nov 28 '18 at 2:04




















                      • I doubt this approach is feasible as I will be solving sixth degree $p(a)< 0$.
                        – henceproved
                        Nov 28 '18 at 1:56












                      • Actually your polynomial always has a double root so the discriminant is always 0 for all $a$ (from which you can work from there). But anyway the above answer is definitely better.
                        – Rcwt
                        Nov 28 '18 at 2:04


















                      I doubt this approach is feasible as I will be solving sixth degree $p(a)< 0$.
                      – henceproved
                      Nov 28 '18 at 1:56






                      I doubt this approach is feasible as I will be solving sixth degree $p(a)< 0$.
                      – henceproved
                      Nov 28 '18 at 1:56














                      Actually your polynomial always has a double root so the discriminant is always 0 for all $a$ (from which you can work from there). But anyway the above answer is definitely better.
                      – Rcwt
                      Nov 28 '18 at 2:04






                      Actually your polynomial always has a double root so the discriminant is always 0 for all $a$ (from which you can work from there). But anyway the above answer is definitely better.
                      – Rcwt
                      Nov 28 '18 at 2:04













                      0














                      Considering that $$p(x) = x^4 + ax^3 + bx^2 + ax + 1$$ If $x=1$ is a root, then
                      $$p(1)=2 a+b+2=0 implies b=-2-2a$$ and
                      $$p(x)=(x-1),(x^3+(a+1) x^2-(a+1) x-1)$$ Clearly, $x=1$ is also a root for the second term making
                      $$p(x)=(x-1)^2,(x^2+(a+2) x+1)$$ So, there are complex roots if the discriminant $Delta$ of the remaining quadratic equation is negative.






                      share|cite|improve this answer


























                        0














                        Considering that $$p(x) = x^4 + ax^3 + bx^2 + ax + 1$$ If $x=1$ is a root, then
                        $$p(1)=2 a+b+2=0 implies b=-2-2a$$ and
                        $$p(x)=(x-1),(x^3+(a+1) x^2-(a+1) x-1)$$ Clearly, $x=1$ is also a root for the second term making
                        $$p(x)=(x-1)^2,(x^2+(a+2) x+1)$$ So, there are complex roots if the discriminant $Delta$ of the remaining quadratic equation is negative.






                        share|cite|improve this answer
























                          0












                          0








                          0






                          Considering that $$p(x) = x^4 + ax^3 + bx^2 + ax + 1$$ If $x=1$ is a root, then
                          $$p(1)=2 a+b+2=0 implies b=-2-2a$$ and
                          $$p(x)=(x-1),(x^3+(a+1) x^2-(a+1) x-1)$$ Clearly, $x=1$ is also a root for the second term making
                          $$p(x)=(x-1)^2,(x^2+(a+2) x+1)$$ So, there are complex roots if the discriminant $Delta$ of the remaining quadratic equation is negative.






                          share|cite|improve this answer












                          Considering that $$p(x) = x^4 + ax^3 + bx^2 + ax + 1$$ If $x=1$ is a root, then
                          $$p(1)=2 a+b+2=0 implies b=-2-2a$$ and
                          $$p(x)=(x-1),(x^3+(a+1) x^2-(a+1) x-1)$$ Clearly, $x=1$ is also a root for the second term making
                          $$p(x)=(x-1)^2,(x^2+(a+2) x+1)$$ So, there are complex roots if the discriminant $Delta$ of the remaining quadratic equation is negative.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 28 '18 at 3:53









                          Claude Leibovici

                          119k1157132




                          119k1157132






























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                              up vote 62 down vote favorite 19 I just watched Stephan T. Lavavej talk at CppCon 2018 on "Class Template Argument Deduction", where at some point he incidentally says: In C++ type information almost never flows backwards ... I had to say "almost" because there's one or two cases, possibly more but very few . Despite trying to figure out which cases he might be referring to, I couldn't come up with anything. Hence the question: In which cases the C++17 standard mandates that type information propagate backwards? c++ language-lawyer c++17 share | improve this question edited 7 hours ago scohe0
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                              Grease: Live!

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                              A tradução deste artigo está abaixo da qualidade média aceitável. É possível que tenha sido feita por um tradutor automático ou por alguém que não conhece bem o português ou a língua original do texto. Caso queira colaborar com a Wikipédia, tente encontrar a página original e melhore este verbete conforme o guia de tradução. Grease: Live! Grease: Live!  (PRT/BRA)  Estados Unidos 2016 •   cor •   130 min. min   Direção Thomas Kail Alex Rudzinski Produção Marc Platt Adam Siegel Roteiro Allan Carr Robert Cary Warren Casey Jim Jacobs Jonathan Tolins Bronte Woodard Baseado em Grease (musical) Grease (filme) Elenco Julianne Hough Aaron Tveit Vanessa Hudgens Carlos Pena Jr. Keke Palmer Carly Rae Jepsen Kether Donohue Jordan Fisher David Del Rio Andrew Call Gênero Musical Música Tom Kitt Figurino William Ivey Long Companhia(s) produtora(s) Warner Bros. Studios Distribuição Fox Lançamento 31 de Janei
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