Relation between roots and coefficients - manipulation of identities












3












$begingroup$



The polynomial $x^3+3x^2-2x+1$ has roots $alpha, beta, gamma$ . Find $$alpha^2(beta + gamma) + beta^2(alpha + gamma) + gamma^2(alpha + beta)$$




I tried finding the relation using $-b/a$, $c/a$ and $-d/a$. I couldn’t seem to find anything. I also tried solving for one root but it gave me back the polynomial but with the root as the variable. Also the polynomial can not be factorised.






polynomials







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    3












    $begingroup$



    The polynomial $x^3+3x^2-2x+1$ has roots $alpha, beta, gamma$ . Find $$alpha^2(beta + gamma) + beta^2(alpha + gamma) + gamma^2(alpha + beta)$$




    I tried finding the relation using $-b/a$, $c/a$ and $-d/a$. I couldn’t seem to find anything. I also tried solving for one root but it gave me back the polynomial but with the root as the variable. Also the polynomial can not be factorised.






    polynomials







    share|cite|improve this question









    New contributor




    Mmloiler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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    $endgroup$















      3












      3








      3





      $begingroup$



      The polynomial $x^3+3x^2-2x+1$ has roots $alpha, beta, gamma$ . Find $$alpha^2(beta + gamma) + beta^2(alpha + gamma) + gamma^2(alpha + beta)$$




      I tried finding the relation using $-b/a$, $c/a$ and $-d/a$. I couldn’t seem to find anything. I also tried solving for one root but it gave me back the polynomial but with the root as the variable. Also the polynomial can not be factorised.






      polynomials







      share|cite|improve this question









      New contributor




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







      $endgroup$





      The polynomial $x^3+3x^2-2x+1$ has roots $alpha, beta, gamma$ . Find $$alpha^2(beta + gamma) + beta^2(alpha + gamma) + gamma^2(alpha + beta)$$




      I tried finding the relation using $-b/a$, $c/a$ and $-d/a$. I couldn’t seem to find anything. I also tried solving for one root but it gave me back the polynomial but with the root as the variable. Also the polynomial can not be factorised.






      polynomials




      polynomials






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      edited 2 hours ago









      Eevee Trainer

      6,41811237




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      asked 2 hours ago









      MmloilerMmloiler

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          3 Answers
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          active

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          4












          $begingroup$

          Since the polynomial has three roots and its highest degree is 3, we can write
          $$
          p(x) = (x-alpha)(x-beta)(x-gamma) = x^3 +3x^2 - 2x + 1.
          $$
          It then follows from
          $$
          x^3 - (alpha+beta+gamma)x^2 + (alphabeta + beta gamma + gamma alpha)x - alphabeta gamma = x^3+3x^2-2x + 1
          $$
          that
          $$
          alpha+beta+gamma = -3, quad alphabeta + beta gamma + gamma alpha = -2, quad
          alphabeta gamma = -1.
          $$
          Note that
          $$
          -2alpha = alpha(alphabeta + beta gamma + gamma alpha) = alpha^2(beta+gamma) +alphabetagamma = alpha^2(beta+gamma) - 1.
          $$
          Thus $alpha^2(beta + gamma)=1-2alpha$.
          Similarly, $beta^2(alpha + gamma) = 1-2beta$ and $gamma^2(alpha + beta) = 1-2gamma$.
          Therefore,
          begin{align}
          alpha^2(beta + gamma) + beta^2(alpha + gamma) + gamma^2(alpha + beta)
          &= (1-2alpha) + (1-2beta) + (1-2gamma) \
          &= 3 -2(alpha+beta+gamma) = 3 +6 =9.
          end{align}






          share|cite|improve this answer









          $endgroup$





















            4












            $begingroup$

            Any symmetric (polynomial) function of the roots can be expressed in terms of the Vieta coefficients. Here, check the hint:
            $$sum alpha^2(beta+gamma) = (alpha+beta+gamma)(alphabeta+betagamma+gammaalpha)-3alphabetagamma$$



            --
            In case you want a systematic method to express in terms of elementary symmetric polynomials, check this answer for Gauss' algorithm.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Another often useful approach is to transform the polynomial (and / or) the result using the given polynomial. For e.g. $sum alpha^2(beta+gamma) = sumalpha^2(-3-alpha) = sum (-alpha^3-3alpha^2) = sum (-2alpha+1)=-2cdot(-3)+3=9 $
              $endgroup$
              – Macavity
              1 hour ago





















            -1












            $begingroup$

            $a,b,c$ are the three roots.
            $$
            begin{align}
            &a^2*(b+c)+b^2*(a+c)+c^2*(a+b)\
            ={}&(a+b+c)*(a^2+b^2+c^2)-(a^3+b^3+c^3)\
            ={}&(-3)*(a^2+b^2+c^2)-(a+b+c)^3\
            ={}&(-3)*((a+b+c)^2-2ab-2ac-2bc)-(a+b+c) * (a^2+b^2+c^2) + ab(a+b) + ac(a+c) + bc(b+c)\
            ={}& (-3)*(9-2*(-2))-(-3)*(9-2*(-2)) + ab(a+b+c-c) + ac(a+b+c-b) + bc(a+b+c-a)\
            ={}&(a+b+c)(ab+ac+bc)-3abc\
            ={}&(-3)*(-2)-3*(-1)\
            ={}&6-(-3)\
            ={}&9
            end{align}
            $$






            share|cite|improve this answer










            New contributor




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






            $endgroup$









            • 5




              $begingroup$
              Welcome to MSE. Note that I, at least, found what you wrote hard to read. To help make your future math formatting look better, I suggest you read & use what it says in MathJax basic tutorial and quick reference.
              $endgroup$
              – John Omielan
              1 hour ago











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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4












            $begingroup$

            Since the polynomial has three roots and its highest degree is 3, we can write
            $$
            p(x) = (x-alpha)(x-beta)(x-gamma) = x^3 +3x^2 - 2x + 1.
            $$
            It then follows from
            $$
            x^3 - (alpha+beta+gamma)x^2 + (alphabeta + beta gamma + gamma alpha)x - alphabeta gamma = x^3+3x^2-2x + 1
            $$
            that
            $$
            alpha+beta+gamma = -3, quad alphabeta + beta gamma + gamma alpha = -2, quad
            alphabeta gamma = -1.
            $$
            Note that
            $$
            -2alpha = alpha(alphabeta + beta gamma + gamma alpha) = alpha^2(beta+gamma) +alphabetagamma = alpha^2(beta+gamma) - 1.
            $$
            Thus $alpha^2(beta + gamma)=1-2alpha$.
            Similarly, $beta^2(alpha + gamma) = 1-2beta$ and $gamma^2(alpha + beta) = 1-2gamma$.
            Therefore,
            begin{align}
            alpha^2(beta + gamma) + beta^2(alpha + gamma) + gamma^2(alpha + beta)
            &= (1-2alpha) + (1-2beta) + (1-2gamma) \
            &= 3 -2(alpha+beta+gamma) = 3 +6 =9.
            end{align}






            share|cite|improve this answer









            $endgroup$


















              4












              $begingroup$

              Since the polynomial has three roots and its highest degree is 3, we can write
              $$
              p(x) = (x-alpha)(x-beta)(x-gamma) = x^3 +3x^2 - 2x + 1.
              $$
              It then follows from
              $$
              x^3 - (alpha+beta+gamma)x^2 + (alphabeta + beta gamma + gamma alpha)x - alphabeta gamma = x^3+3x^2-2x + 1
              $$
              that
              $$
              alpha+beta+gamma = -3, quad alphabeta + beta gamma + gamma alpha = -2, quad
              alphabeta gamma = -1.
              $$
              Note that
              $$
              -2alpha = alpha(alphabeta + beta gamma + gamma alpha) = alpha^2(beta+gamma) +alphabetagamma = alpha^2(beta+gamma) - 1.
              $$
              Thus $alpha^2(beta + gamma)=1-2alpha$.
              Similarly, $beta^2(alpha + gamma) = 1-2beta$ and $gamma^2(alpha + beta) = 1-2gamma$.
              Therefore,
              begin{align}
              alpha^2(beta + gamma) + beta^2(alpha + gamma) + gamma^2(alpha + beta)
              &= (1-2alpha) + (1-2beta) + (1-2gamma) \
              &= 3 -2(alpha+beta+gamma) = 3 +6 =9.
              end{align}






              share|cite|improve this answer









              $endgroup$
















                4












                4








                4





                $begingroup$

                Since the polynomial has three roots and its highest degree is 3, we can write
                $$
                p(x) = (x-alpha)(x-beta)(x-gamma) = x^3 +3x^2 - 2x + 1.
                $$
                It then follows from
                $$
                x^3 - (alpha+beta+gamma)x^2 + (alphabeta + beta gamma + gamma alpha)x - alphabeta gamma = x^3+3x^2-2x + 1
                $$
                that
                $$
                alpha+beta+gamma = -3, quad alphabeta + beta gamma + gamma alpha = -2, quad
                alphabeta gamma = -1.
                $$
                Note that
                $$
                -2alpha = alpha(alphabeta + beta gamma + gamma alpha) = alpha^2(beta+gamma) +alphabetagamma = alpha^2(beta+gamma) - 1.
                $$
                Thus $alpha^2(beta + gamma)=1-2alpha$.
                Similarly, $beta^2(alpha + gamma) = 1-2beta$ and $gamma^2(alpha + beta) = 1-2gamma$.
                Therefore,
                begin{align}
                alpha^2(beta + gamma) + beta^2(alpha + gamma) + gamma^2(alpha + beta)
                &= (1-2alpha) + (1-2beta) + (1-2gamma) \
                &= 3 -2(alpha+beta+gamma) = 3 +6 =9.
                end{align}






                share|cite|improve this answer









                $endgroup$



                Since the polynomial has three roots and its highest degree is 3, we can write
                $$
                p(x) = (x-alpha)(x-beta)(x-gamma) = x^3 +3x^2 - 2x + 1.
                $$
                It then follows from
                $$
                x^3 - (alpha+beta+gamma)x^2 + (alphabeta + beta gamma + gamma alpha)x - alphabeta gamma = x^3+3x^2-2x + 1
                $$
                that
                $$
                alpha+beta+gamma = -3, quad alphabeta + beta gamma + gamma alpha = -2, quad
                alphabeta gamma = -1.
                $$
                Note that
                $$
                -2alpha = alpha(alphabeta + beta gamma + gamma alpha) = alpha^2(beta+gamma) +alphabetagamma = alpha^2(beta+gamma) - 1.
                $$
                Thus $alpha^2(beta + gamma)=1-2alpha$.
                Similarly, $beta^2(alpha + gamma) = 1-2beta$ and $gamma^2(alpha + beta) = 1-2gamma$.
                Therefore,
                begin{align}
                alpha^2(beta + gamma) + beta^2(alpha + gamma) + gamma^2(alpha + beta)
                &= (1-2alpha) + (1-2beta) + (1-2gamma) \
                &= 3 -2(alpha+beta+gamma) = 3 +6 =9.
                end{align}







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 hours ago









                induction601induction601

                1,226314




                1,226314























                    4












                    $begingroup$

                    Any symmetric (polynomial) function of the roots can be expressed in terms of the Vieta coefficients. Here, check the hint:
                    $$sum alpha^2(beta+gamma) = (alpha+beta+gamma)(alphabeta+betagamma+gammaalpha)-3alphabetagamma$$



                    --
                    In case you want a systematic method to express in terms of elementary symmetric polynomials, check this answer for Gauss' algorithm.






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      Another often useful approach is to transform the polynomial (and / or) the result using the given polynomial. For e.g. $sum alpha^2(beta+gamma) = sumalpha^2(-3-alpha) = sum (-alpha^3-3alpha^2) = sum (-2alpha+1)=-2cdot(-3)+3=9 $
                      $endgroup$
                      – Macavity
                      1 hour ago


















                    4












                    $begingroup$

                    Any symmetric (polynomial) function of the roots can be expressed in terms of the Vieta coefficients. Here, check the hint:
                    $$sum alpha^2(beta+gamma) = (alpha+beta+gamma)(alphabeta+betagamma+gammaalpha)-3alphabetagamma$$



                    --
                    In case you want a systematic method to express in terms of elementary symmetric polynomials, check this answer for Gauss' algorithm.






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      Another often useful approach is to transform the polynomial (and / or) the result using the given polynomial. For e.g. $sum alpha^2(beta+gamma) = sumalpha^2(-3-alpha) = sum (-alpha^3-3alpha^2) = sum (-2alpha+1)=-2cdot(-3)+3=9 $
                      $endgroup$
                      – Macavity
                      1 hour ago
















                    4












                    4








                    4





                    $begingroup$

                    Any symmetric (polynomial) function of the roots can be expressed in terms of the Vieta coefficients. Here, check the hint:
                    $$sum alpha^2(beta+gamma) = (alpha+beta+gamma)(alphabeta+betagamma+gammaalpha)-3alphabetagamma$$



                    --
                    In case you want a systematic method to express in terms of elementary symmetric polynomials, check this answer for Gauss' algorithm.






                    share|cite|improve this answer











                    $endgroup$



                    Any symmetric (polynomial) function of the roots can be expressed in terms of the Vieta coefficients. Here, check the hint:
                    $$sum alpha^2(beta+gamma) = (alpha+beta+gamma)(alphabeta+betagamma+gammaalpha)-3alphabetagamma$$



                    --
                    In case you want a systematic method to express in terms of elementary symmetric polynomials, check this answer for Gauss' algorithm.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 1 hour ago

























                    answered 2 hours ago









                    MacavityMacavity

                    35.5k52554




                    35.5k52554












                    • $begingroup$
                      Another often useful approach is to transform the polynomial (and / or) the result using the given polynomial. For e.g. $sum alpha^2(beta+gamma) = sumalpha^2(-3-alpha) = sum (-alpha^3-3alpha^2) = sum (-2alpha+1)=-2cdot(-3)+3=9 $
                      $endgroup$
                      – Macavity
                      1 hour ago




















                    • $begingroup$
                      Another often useful approach is to transform the polynomial (and / or) the result using the given polynomial. For e.g. $sum alpha^2(beta+gamma) = sumalpha^2(-3-alpha) = sum (-alpha^3-3alpha^2) = sum (-2alpha+1)=-2cdot(-3)+3=9 $
                      $endgroup$
                      – Macavity
                      1 hour ago


















                    $begingroup$
                    Another often useful approach is to transform the polynomial (and / or) the result using the given polynomial. For e.g. $sum alpha^2(beta+gamma) = sumalpha^2(-3-alpha) = sum (-alpha^3-3alpha^2) = sum (-2alpha+1)=-2cdot(-3)+3=9 $
                    $endgroup$
                    – Macavity
                    1 hour ago






                    $begingroup$
                    Another often useful approach is to transform the polynomial (and / or) the result using the given polynomial. For e.g. $sum alpha^2(beta+gamma) = sumalpha^2(-3-alpha) = sum (-alpha^3-3alpha^2) = sum (-2alpha+1)=-2cdot(-3)+3=9 $
                    $endgroup$
                    – Macavity
                    1 hour ago













                    -1












                    $begingroup$

                    $a,b,c$ are the three roots.
                    $$
                    begin{align}
                    &a^2*(b+c)+b^2*(a+c)+c^2*(a+b)\
                    ={}&(a+b+c)*(a^2+b^2+c^2)-(a^3+b^3+c^3)\
                    ={}&(-3)*(a^2+b^2+c^2)-(a+b+c)^3\
                    ={}&(-3)*((a+b+c)^2-2ab-2ac-2bc)-(a+b+c) * (a^2+b^2+c^2) + ab(a+b) + ac(a+c) + bc(b+c)\
                    ={}& (-3)*(9-2*(-2))-(-3)*(9-2*(-2)) + ab(a+b+c-c) + ac(a+b+c-b) + bc(a+b+c-a)\
                    ={}&(a+b+c)(ab+ac+bc)-3abc\
                    ={}&(-3)*(-2)-3*(-1)\
                    ={}&6-(-3)\
                    ={}&9
                    end{align}
                    $$






                    share|cite|improve this answer










                    New contributor




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






                    $endgroup$









                    • 5




                      $begingroup$
                      Welcome to MSE. Note that I, at least, found what you wrote hard to read. To help make your future math formatting look better, I suggest you read & use what it says in MathJax basic tutorial and quick reference.
                      $endgroup$
                      – John Omielan
                      1 hour ago
















                    -1












                    $begingroup$

                    $a,b,c$ are the three roots.
                    $$
                    begin{align}
                    &a^2*(b+c)+b^2*(a+c)+c^2*(a+b)\
                    ={}&(a+b+c)*(a^2+b^2+c^2)-(a^3+b^3+c^3)\
                    ={}&(-3)*(a^2+b^2+c^2)-(a+b+c)^3\
                    ={}&(-3)*((a+b+c)^2-2ab-2ac-2bc)-(a+b+c) * (a^2+b^2+c^2) + ab(a+b) + ac(a+c) + bc(b+c)\
                    ={}& (-3)*(9-2*(-2))-(-3)*(9-2*(-2)) + ab(a+b+c-c) + ac(a+b+c-b) + bc(a+b+c-a)\
                    ={}&(a+b+c)(ab+ac+bc)-3abc\
                    ={}&(-3)*(-2)-3*(-1)\
                    ={}&6-(-3)\
                    ={}&9
                    end{align}
                    $$






                    share|cite|improve this answer










                    New contributor




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






                    $endgroup$









                    • 5




                      $begingroup$
                      Welcome to MSE. Note that I, at least, found what you wrote hard to read. To help make your future math formatting look better, I suggest you read & use what it says in MathJax basic tutorial and quick reference.
                      $endgroup$
                      – John Omielan
                      1 hour ago














                    -1












                    -1








                    -1





                    $begingroup$

                    $a,b,c$ are the three roots.
                    $$
                    begin{align}
                    &a^2*(b+c)+b^2*(a+c)+c^2*(a+b)\
                    ={}&(a+b+c)*(a^2+b^2+c^2)-(a^3+b^3+c^3)\
                    ={}&(-3)*(a^2+b^2+c^2)-(a+b+c)^3\
                    ={}&(-3)*((a+b+c)^2-2ab-2ac-2bc)-(a+b+c) * (a^2+b^2+c^2) + ab(a+b) + ac(a+c) + bc(b+c)\
                    ={}& (-3)*(9-2*(-2))-(-3)*(9-2*(-2)) + ab(a+b+c-c) + ac(a+b+c-b) + bc(a+b+c-a)\
                    ={}&(a+b+c)(ab+ac+bc)-3abc\
                    ={}&(-3)*(-2)-3*(-1)\
                    ={}&6-(-3)\
                    ={}&9
                    end{align}
                    $$






                    share|cite|improve this answer










                    New contributor




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






                    $endgroup$



                    $a,b,c$ are the three roots.
                    $$
                    begin{align}
                    &a^2*(b+c)+b^2*(a+c)+c^2*(a+b)\
                    ={}&(a+b+c)*(a^2+b^2+c^2)-(a^3+b^3+c^3)\
                    ={}&(-3)*(a^2+b^2+c^2)-(a+b+c)^3\
                    ={}&(-3)*((a+b+c)^2-2ab-2ac-2bc)-(a+b+c) * (a^2+b^2+c^2) + ab(a+b) + ac(a+c) + bc(b+c)\
                    ={}& (-3)*(9-2*(-2))-(-3)*(9-2*(-2)) + ab(a+b+c-c) + ac(a+b+c-b) + bc(a+b+c-a)\
                    ={}&(a+b+c)(ab+ac+bc)-3abc\
                    ={}&(-3)*(-2)-3*(-1)\
                    ={}&6-(-3)\
                    ={}&9
                    end{align}
                    $$







                    share|cite|improve this answer










                    New contributor




                    Wentao Wang 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 answer



                    share|cite|improve this answer








                    edited 39 mins ago









                    Brahadeesh

                    6,46942363




                    6,46942363






                    New contributor




                    Wentao Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                    answered 2 hours ago









                    Wentao WangWentao Wang

                    11




                    11




                    New contributor




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





                    New contributor





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






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








                    • 5




                      $begingroup$
                      Welcome to MSE. Note that I, at least, found what you wrote hard to read. To help make your future math formatting look better, I suggest you read & use what it says in MathJax basic tutorial and quick reference.
                      $endgroup$
                      – John Omielan
                      1 hour ago














                    • 5




                      $begingroup$
                      Welcome to MSE. Note that I, at least, found what you wrote hard to read. To help make your future math formatting look better, I suggest you read & use what it says in MathJax basic tutorial and quick reference.
                      $endgroup$
                      – John Omielan
                      1 hour ago








                    5




                    5




                    $begingroup$
                    Welcome to MSE. Note that I, at least, found what you wrote hard to read. To help make your future math formatting look better, I suggest you read & use what it says in MathJax basic tutorial and quick reference.
                    $endgroup$
                    – John Omielan
                    1 hour ago




                    $begingroup$
                    Welcome to MSE. Note that I, at least, found what you wrote hard to read. To help make your future math formatting look better, I suggest you read & use what it says in MathJax basic tutorial and quick reference.
                    $endgroup$
                    – John Omielan
                    1 hour ago










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