If $a$, $b$ are the roots of $x^2-2x+3$.Then the equation whose roots are $a^3-3a^2+5a-2$ and $b^3-b^2+b+5$...












4












$begingroup$



If $a$, $b$ are the roots of $x^2-2x+3$.Then the equation whose roots are $a^3-3a^2+5a-2$ and $b^3-b^2+b+5$ is:




I have not been able to find a better method than to calculate $a$ and $b$ then substitute them into the roots for the new polynomial.



I believe this question can't be transformed in a similar manner as mentioned in this question as the new roots are asymmetrical.



Does a better method than the lackluster substitution, exist?



The answer is:




$x^2-3x+2$







linear-algebra







share|cite|improve this question









$endgroup$

















    4












    $begingroup$



    If $a$, $b$ are the roots of $x^2-2x+3$.Then the equation whose roots are $a^3-3a^2+5a-2$ and $b^3-b^2+b+5$ is:




    I have not been able to find a better method than to calculate $a$ and $b$ then substitute them into the roots for the new polynomial.



    I believe this question can't be transformed in a similar manner as mentioned in this question as the new roots are asymmetrical.



    Does a better method than the lackluster substitution, exist?



    The answer is:




    $x^2-3x+2$







    linear-algebra







    share|cite|improve this question









    $endgroup$















      4












      4








      4


      2



      $begingroup$



      If $a$, $b$ are the roots of $x^2-2x+3$.Then the equation whose roots are $a^3-3a^2+5a-2$ and $b^3-b^2+b+5$ is:




      I have not been able to find a better method than to calculate $a$ and $b$ then substitute them into the roots for the new polynomial.



      I believe this question can't be transformed in a similar manner as mentioned in this question as the new roots are asymmetrical.



      Does a better method than the lackluster substitution, exist?



      The answer is:




      $x^2-3x+2$







      linear-algebra







      share|cite|improve this question









      $endgroup$





      If $a$, $b$ are the roots of $x^2-2x+3$.Then the equation whose roots are $a^3-3a^2+5a-2$ and $b^3-b^2+b+5$ is:




      I have not been able to find a better method than to calculate $a$ and $b$ then substitute them into the roots for the new polynomial.



      I believe this question can't be transformed in a similar manner as mentioned in this question as the new roots are asymmetrical.



      Does a better method than the lackluster substitution, exist?



      The answer is:




      $x^2-3x+2$







      linear-algebra




      linear-algebra






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 24 '18 at 16:12









      CaptainQuestionCaptainQuestion

      1337




      1337






















          3 Answers
          3






          active

          oldest

          votes


















          5












          $begingroup$

          Since $x^2=2x-3$, we get that $x^3=2x^2-3x=x-6$. Then,
          $$
          begin{align}
          a^3-3a^2+5a-2
          &=(a-6)-3(2a-3)+5a-2\
          &=1
          end{align}
          $$
          and
          $$
          begin{align}
          b^3-b^2+b+5
          &=(b-6)-(2b-3)+b+5\
          &=2
          end{align}
          $$
          It is easy to find an equation which has roots of $1$ and $2$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Never thought about using the initial polynomial to create a reduction formula absolutely radical!
            $endgroup$
            – CaptainQuestion
            Dec 24 '18 at 16:43






          • 1




            $begingroup$
            @CaptainQuestion In this case it's actually easier to use long division to compute the remainder - I added that to my answer to show how easy it is.
            $endgroup$
            – Bill Dubuque
            Dec 24 '18 at 16:44





















          5












          $begingroup$

          Hint:



          As $a,b$ are the roots of $x^2-2x+3=0$



          $a^3-3a^2+5a-3=(a^2-2a+3)(a-1)+1=1$



          Similarly for $b$






          share|cite|improve this answer









          $endgroup$





















            4












            $begingroup$

            Hint $ x^3-3x^2+5x-2,bmod, color{#c00}{x^2-2x+3}, =, color{#0a0}1 $ (and $= color{#90f}2$ for the other). So we seek a polynomial with roots $color{#0a0}1$ and $color{#90f}2,,$ e.g. $ (x-color{#0a0}1)(x-color{#90f}2)$



            Remark $ $ The remainder is quickly computable by long division (ignoring the unneeded quotient)



            $$begin{align}
            & 1 {-}3 , 5, {-}2\
            &color{#c00}{{-}1, 2 {-}3}\
            & {-}1 2 {-}2\
            &color{#c00}{ 1 {-}2 3}\
            &qquadqquadquad color{#0a0}1
            end{align}qquadqquadquad$$






            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

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              5












              $begingroup$

              Since $x^2=2x-3$, we get that $x^3=2x^2-3x=x-6$. Then,
              $$
              begin{align}
              a^3-3a^2+5a-2
              &=(a-6)-3(2a-3)+5a-2\
              &=1
              end{align}
              $$
              and
              $$
              begin{align}
              b^3-b^2+b+5
              &=(b-6)-(2b-3)+b+5\
              &=2
              end{align}
              $$
              It is easy to find an equation which has roots of $1$ and $2$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Never thought about using the initial polynomial to create a reduction formula absolutely radical!
                $endgroup$
                – CaptainQuestion
                Dec 24 '18 at 16:43






              • 1




                $begingroup$
                @CaptainQuestion In this case it's actually easier to use long division to compute the remainder - I added that to my answer to show how easy it is.
                $endgroup$
                – Bill Dubuque
                Dec 24 '18 at 16:44


















              5












              $begingroup$

              Since $x^2=2x-3$, we get that $x^3=2x^2-3x=x-6$. Then,
              $$
              begin{align}
              a^3-3a^2+5a-2
              &=(a-6)-3(2a-3)+5a-2\
              &=1
              end{align}
              $$
              and
              $$
              begin{align}
              b^3-b^2+b+5
              &=(b-6)-(2b-3)+b+5\
              &=2
              end{align}
              $$
              It is easy to find an equation which has roots of $1$ and $2$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Never thought about using the initial polynomial to create a reduction formula absolutely radical!
                $endgroup$
                – CaptainQuestion
                Dec 24 '18 at 16:43






              • 1




                $begingroup$
                @CaptainQuestion In this case it's actually easier to use long division to compute the remainder - I added that to my answer to show how easy it is.
                $endgroup$
                – Bill Dubuque
                Dec 24 '18 at 16:44
















              5












              5








              5





              $begingroup$

              Since $x^2=2x-3$, we get that $x^3=2x^2-3x=x-6$. Then,
              $$
              begin{align}
              a^3-3a^2+5a-2
              &=(a-6)-3(2a-3)+5a-2\
              &=1
              end{align}
              $$
              and
              $$
              begin{align}
              b^3-b^2+b+5
              &=(b-6)-(2b-3)+b+5\
              &=2
              end{align}
              $$
              It is easy to find an equation which has roots of $1$ and $2$.






              share|cite|improve this answer









              $endgroup$



              Since $x^2=2x-3$, we get that $x^3=2x^2-3x=x-6$. Then,
              $$
              begin{align}
              a^3-3a^2+5a-2
              &=(a-6)-3(2a-3)+5a-2\
              &=1
              end{align}
              $$
              and
              $$
              begin{align}
              b^3-b^2+b+5
              &=(b-6)-(2b-3)+b+5\
              &=2
              end{align}
              $$
              It is easy to find an equation which has roots of $1$ and $2$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 24 '18 at 16:29









              robjohn♦robjohn

              269k27311638




              269k27311638












              • $begingroup$
                Never thought about using the initial polynomial to create a reduction formula absolutely radical!
                $endgroup$
                – CaptainQuestion
                Dec 24 '18 at 16:43






              • 1




                $begingroup$
                @CaptainQuestion In this case it's actually easier to use long division to compute the remainder - I added that to my answer to show how easy it is.
                $endgroup$
                – Bill Dubuque
                Dec 24 '18 at 16:44




















              • $begingroup$
                Never thought about using the initial polynomial to create a reduction formula absolutely radical!
                $endgroup$
                – CaptainQuestion
                Dec 24 '18 at 16:43






              • 1




                $begingroup$
                @CaptainQuestion In this case it's actually easier to use long division to compute the remainder - I added that to my answer to show how easy it is.
                $endgroup$
                – Bill Dubuque
                Dec 24 '18 at 16:44


















              $begingroup$
              Never thought about using the initial polynomial to create a reduction formula absolutely radical!
              $endgroup$
              – CaptainQuestion
              Dec 24 '18 at 16:43




              $begingroup$
              Never thought about using the initial polynomial to create a reduction formula absolutely radical!
              $endgroup$
              – CaptainQuestion
              Dec 24 '18 at 16:43




              1




              1




              $begingroup$
              @CaptainQuestion In this case it's actually easier to use long division to compute the remainder - I added that to my answer to show how easy it is.
              $endgroup$
              – Bill Dubuque
              Dec 24 '18 at 16:44






              $begingroup$
              @CaptainQuestion In this case it's actually easier to use long division to compute the remainder - I added that to my answer to show how easy it is.
              $endgroup$
              – Bill Dubuque
              Dec 24 '18 at 16:44













              5












              $begingroup$

              Hint:



              As $a,b$ are the roots of $x^2-2x+3=0$



              $a^3-3a^2+5a-3=(a^2-2a+3)(a-1)+1=1$



              Similarly for $b$






              share|cite|improve this answer









              $endgroup$


















                5












                $begingroup$

                Hint:



                As $a,b$ are the roots of $x^2-2x+3=0$



                $a^3-3a^2+5a-3=(a^2-2a+3)(a-1)+1=1$



                Similarly for $b$






                share|cite|improve this answer









                $endgroup$
















                  5












                  5








                  5





                  $begingroup$

                  Hint:



                  As $a,b$ are the roots of $x^2-2x+3=0$



                  $a^3-3a^2+5a-3=(a^2-2a+3)(a-1)+1=1$



                  Similarly for $b$






                  share|cite|improve this answer









                  $endgroup$



                  Hint:



                  As $a,b$ are the roots of $x^2-2x+3=0$



                  $a^3-3a^2+5a-3=(a^2-2a+3)(a-1)+1=1$



                  Similarly for $b$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 24 '18 at 16:20









                  lab bhattacharjeelab bhattacharjee

                  226k15158275




                  226k15158275























                      4












                      $begingroup$

                      Hint $ x^3-3x^2+5x-2,bmod, color{#c00}{x^2-2x+3}, =, color{#0a0}1 $ (and $= color{#90f}2$ for the other). So we seek a polynomial with roots $color{#0a0}1$ and $color{#90f}2,,$ e.g. $ (x-color{#0a0}1)(x-color{#90f}2)$



                      Remark $ $ The remainder is quickly computable by long division (ignoring the unneeded quotient)



                      $$begin{align}
                      & 1 {-}3 , 5, {-}2\
                      &color{#c00}{{-}1, 2 {-}3}\
                      & {-}1 2 {-}2\
                      &color{#c00}{ 1 {-}2 3}\
                      &qquadqquadquad color{#0a0}1
                      end{align}qquadqquadquad$$






                      share|cite|improve this answer











                      $endgroup$


















                        4












                        $begingroup$

                        Hint $ x^3-3x^2+5x-2,bmod, color{#c00}{x^2-2x+3}, =, color{#0a0}1 $ (and $= color{#90f}2$ for the other). So we seek a polynomial with roots $color{#0a0}1$ and $color{#90f}2,,$ e.g. $ (x-color{#0a0}1)(x-color{#90f}2)$



                        Remark $ $ The remainder is quickly computable by long division (ignoring the unneeded quotient)



                        $$begin{align}
                        & 1 {-}3 , 5, {-}2\
                        &color{#c00}{{-}1, 2 {-}3}\
                        & {-}1 2 {-}2\
                        &color{#c00}{ 1 {-}2 3}\
                        &qquadqquadquad color{#0a0}1
                        end{align}qquadqquadquad$$






                        share|cite|improve this answer











                        $endgroup$
















                          4












                          4








                          4





                          $begingroup$

                          Hint $ x^3-3x^2+5x-2,bmod, color{#c00}{x^2-2x+3}, =, color{#0a0}1 $ (and $= color{#90f}2$ for the other). So we seek a polynomial with roots $color{#0a0}1$ and $color{#90f}2,,$ e.g. $ (x-color{#0a0}1)(x-color{#90f}2)$



                          Remark $ $ The remainder is quickly computable by long division (ignoring the unneeded quotient)



                          $$begin{align}
                          & 1 {-}3 , 5, {-}2\
                          &color{#c00}{{-}1, 2 {-}3}\
                          & {-}1 2 {-}2\
                          &color{#c00}{ 1 {-}2 3}\
                          &qquadqquadquad color{#0a0}1
                          end{align}qquadqquadquad$$






                          share|cite|improve this answer











                          $endgroup$



                          Hint $ x^3-3x^2+5x-2,bmod, color{#c00}{x^2-2x+3}, =, color{#0a0}1 $ (and $= color{#90f}2$ for the other). So we seek a polynomial with roots $color{#0a0}1$ and $color{#90f}2,,$ e.g. $ (x-color{#0a0}1)(x-color{#90f}2)$



                          Remark $ $ The remainder is quickly computable by long division (ignoring the unneeded quotient)



                          $$begin{align}
                          & 1 {-}3 , 5, {-}2\
                          &color{#c00}{{-}1, 2 {-}3}\
                          & {-}1 2 {-}2\
                          &color{#c00}{ 1 {-}2 3}\
                          &qquadqquadquad color{#0a0}1
                          end{align}qquadqquadquad$$







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Dec 24 '18 at 16:44

























                          answered Dec 24 '18 at 16:19









                          Bill DubuqueBill Dubuque

                          212k29195651




                          212k29195651






























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