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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                              Aardman Animations

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                              Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
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                              Are they similar matrix

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                              0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
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