Find the remainder of the division of $x^n+5$ with $x^3+10x^2+25x$












1












$begingroup$


Find the remainder of the division of $x^n+5$ with $x^3+10x^2+25x$ over $mathbb{Q}$



What I tried to do is to write $x^n+5=p(x)(x^3+10x^2+25x)+Ax^2+Bx+C$, where $p(x)$ is a polynomial of degree $n-3$. If I set $x=0$ I obtain that $C=5$.



Now $x^3+10x^2+25x=x(x+5)^2$ and by setting $x=-5$, I get that



$5A-B=(-1)^n5^{n-1}$



But I need one more equation to be able to find the coefficients of the remainder, and I can't get one. What should I do?






algebra-precalculus







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$endgroup$








  • 1




    $begingroup$
    Don't you get $C=5$ by setting $x=0$?
    $endgroup$
    – Alejandro Nasif Salum
    Jan 29 at 19:54








  • 1




    $begingroup$
    Hint: differentiate, since $,x = -5,$ is a double root you can again evaluate at $,x = -5,$ to get another equation in $A,B$
    $endgroup$
    – Bill Dubuque
    Jan 29 at 19:58


















1












$begingroup$


Find the remainder of the division of $x^n+5$ with $x^3+10x^2+25x$ over $mathbb{Q}$



What I tried to do is to write $x^n+5=p(x)(x^3+10x^2+25x)+Ax^2+Bx+C$, where $p(x)$ is a polynomial of degree $n-3$. If I set $x=0$ I obtain that $C=5$.



Now $x^3+10x^2+25x=x(x+5)^2$ and by setting $x=-5$, I get that



$5A-B=(-1)^n5^{n-1}$



But I need one more equation to be able to find the coefficients of the remainder, and I can't get one. What should I do?






algebra-precalculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Don't you get $C=5$ by setting $x=0$?
    $endgroup$
    – Alejandro Nasif Salum
    Jan 29 at 19:54








  • 1




    $begingroup$
    Hint: differentiate, since $,x = -5,$ is a double root you can again evaluate at $,x = -5,$ to get another equation in $A,B$
    $endgroup$
    – Bill Dubuque
    Jan 29 at 19:58
















1












1








1





$begingroup$


Find the remainder of the division of $x^n+5$ with $x^3+10x^2+25x$ over $mathbb{Q}$



What I tried to do is to write $x^n+5=p(x)(x^3+10x^2+25x)+Ax^2+Bx+C$, where $p(x)$ is a polynomial of degree $n-3$. If I set $x=0$ I obtain that $C=5$.



Now $x^3+10x^2+25x=x(x+5)^2$ and by setting $x=-5$, I get that



$5A-B=(-1)^n5^{n-1}$



But I need one more equation to be able to find the coefficients of the remainder, and I can't get one. What should I do?






algebra-precalculus







share|cite|improve this question











$endgroup$




Find the remainder of the division of $x^n+5$ with $x^3+10x^2+25x$ over $mathbb{Q}$



What I tried to do is to write $x^n+5=p(x)(x^3+10x^2+25x)+Ax^2+Bx+C$, where $p(x)$ is a polynomial of degree $n-3$. If I set $x=0$ I obtain that $C=5$.



Now $x^3+10x^2+25x=x(x+5)^2$ and by setting $x=-5$, I get that



$5A-B=(-1)^n5^{n-1}$



But I need one more equation to be able to find the coefficients of the remainder, and I can't get one. What should I do?






algebra-precalculus




algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 29 at 20:21







user3701033

















asked Jan 29 at 19:51









user3701033user3701033

1519




1519








  • 1




    $begingroup$
    Don't you get $C=5$ by setting $x=0$?
    $endgroup$
    – Alejandro Nasif Salum
    Jan 29 at 19:54








  • 1




    $begingroup$
    Hint: differentiate, since $,x = -5,$ is a double root you can again evaluate at $,x = -5,$ to get another equation in $A,B$
    $endgroup$
    – Bill Dubuque
    Jan 29 at 19:58
















  • 1




    $begingroup$
    Don't you get $C=5$ by setting $x=0$?
    $endgroup$
    – Alejandro Nasif Salum
    Jan 29 at 19:54








  • 1




    $begingroup$
    Hint: differentiate, since $,x = -5,$ is a double root you can again evaluate at $,x = -5,$ to get another equation in $A,B$
    $endgroup$
    – Bill Dubuque
    Jan 29 at 19:58










1




1




$begingroup$
Don't you get $C=5$ by setting $x=0$?
$endgroup$
– Alejandro Nasif Salum
Jan 29 at 19:54






$begingroup$
Don't you get $C=5$ by setting $x=0$?
$endgroup$
– Alejandro Nasif Salum
Jan 29 at 19:54






1




1




$begingroup$
Hint: differentiate, since $,x = -5,$ is a double root you can again evaluate at $,x = -5,$ to get another equation in $A,B$
$endgroup$
– Bill Dubuque
Jan 29 at 19:58






$begingroup$
Hint: differentiate, since $,x = -5,$ is a double root you can again evaluate at $,x = -5,$ to get another equation in $A,B$
$endgroup$
– Bill Dubuque
Jan 29 at 19:58












3 Answers
3






active

oldest

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2












$begingroup$

It is a good idea to use the roots of $q(x)=x^3+10x^2+25x$, but yes: because $-5$ is a double root, things get a little bit trickier. (Also, check that setting $x=0$ gives $C=5$, not $C=0$).



HINT: But since $-5$ is a double root of $q$, it is both a root of $q$ and $q'$. So try to take the derivative at both members and then take again $x=-5$.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Hint $ $ Subtracting $, C = 5,$ then cancelling $x$ yields



    $qquad f(x) := x^{large n-1} = B + Ax + (x+5)^{large 2} p(x) $



    Thus $, B + Ax = f(-5) + f'(-5)(x+5),$ by Taylor expansion at $,x = -5,$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      If you don't know (or can't use) Taylor then you could instead apply this (purely algebraic) Polynomial Double Root Test. on $,x^{n-1}!-B-Ax,,$ or just differentiate as I hinted on the question.
      $endgroup$
      – Bill Dubuque
      Jan 29 at 20:22





















    0












    $begingroup$

    I think there is an error in the attempt that is part of the question. Anyway, here are some relevant quotients and remainders:



    $$ left( x^{4} + 5 right) = left( x^{3} + 10 x^{2} + 25 x right) cdot left( x - 10 right) + left( 75 x^{2} + 250 x + 5 right) $$
    $$ left( x^{5} + 5 right) = left( x^{3} + 10 x^{2} + 25 x right) cdot left( x^{2} - 10 x + 75 right) + left( - 500 x^{2} - 1875 x + 5 right) $$
    $$ left( x^{6} + 5 right) = left( x^{3} + 10 x^{2} + 25 x right) cdot left( x^{3} - 10 x^{2} + 75 x - 500 right) + left( 3125 x^{2} + 12500 x + 5 right) $$
    $$ left( x^{7} + 5 right) = left( x^{3} + 10 x^{2} + 25 x right) cdot left( x^{4} - 10 x^{3} + 75 x^{2} - 500 x + 3125 right) + left( - 18750 x^{2} - 78125 x + 5 right) $$
    $$ left( x^{8} + 5 right) = left( x^{3} + 10 x^{2} + 25 x right) cdot left( x^{5} - 10 x^{4} + 75 x^{3} - 500 x^{2} + 3125 x - 18750 right) + left( 109375 x^{2} + 468750 x + 5 right) $$
    $$ left( x^{9} + 5 right) = left( x^{3} + 10 x^{2} + 25 x right) cdot left( x^{6} - 10 x^{5} + 75 x^{4} - 500 x^{3} + 3125 x^{2} - 18750 x + 109375 right) + left( - 625000 x^{2} - 2734375 x + 5 right) $$
    $$ left( x^{10} + 5 right) = left( x^{3} + 10 x^{2} + 25 x right) cdot left( x^{7} - 10 x^{6} + 75 x^{5} - 500 x^{4} + 3125 x^{3} - 18750 x^{2} + 109375 x - 625000 right) + left( 3515625 x^{2} + 15625000 x + 5 right) $$






    share|cite|improve this answer











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      3 Answers
      3






      active

      oldest

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      3 Answers
      3






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      2












      $begingroup$

      It is a good idea to use the roots of $q(x)=x^3+10x^2+25x$, but yes: because $-5$ is a double root, things get a little bit trickier. (Also, check that setting $x=0$ gives $C=5$, not $C=0$).



      HINT: But since $-5$ is a double root of $q$, it is both a root of $q$ and $q'$. So try to take the derivative at both members and then take again $x=-5$.






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        It is a good idea to use the roots of $q(x)=x^3+10x^2+25x$, but yes: because $-5$ is a double root, things get a little bit trickier. (Also, check that setting $x=0$ gives $C=5$, not $C=0$).



        HINT: But since $-5$ is a double root of $q$, it is both a root of $q$ and $q'$. So try to take the derivative at both members and then take again $x=-5$.






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          It is a good idea to use the roots of $q(x)=x^3+10x^2+25x$, but yes: because $-5$ is a double root, things get a little bit trickier. (Also, check that setting $x=0$ gives $C=5$, not $C=0$).



          HINT: But since $-5$ is a double root of $q$, it is both a root of $q$ and $q'$. So try to take the derivative at both members and then take again $x=-5$.






          share|cite|improve this answer











          $endgroup$



          It is a good idea to use the roots of $q(x)=x^3+10x^2+25x$, but yes: because $-5$ is a double root, things get a little bit trickier. (Also, check that setting $x=0$ gives $C=5$, not $C=0$).



          HINT: But since $-5$ is a double root of $q$, it is both a root of $q$ and $q'$. So try to take the derivative at both members and then take again $x=-5$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 29 at 20:31

























          answered Jan 29 at 20:00









          Alejandro Nasif SalumAlejandro Nasif Salum

          4,765118




          4,765118























              1












              $begingroup$

              Hint $ $ Subtracting $, C = 5,$ then cancelling $x$ yields



              $qquad f(x) := x^{large n-1} = B + Ax + (x+5)^{large 2} p(x) $



              Thus $, B + Ax = f(-5) + f'(-5)(x+5),$ by Taylor expansion at $,x = -5,$






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                If you don't know (or can't use) Taylor then you could instead apply this (purely algebraic) Polynomial Double Root Test. on $,x^{n-1}!-B-Ax,,$ or just differentiate as I hinted on the question.
                $endgroup$
                – Bill Dubuque
                Jan 29 at 20:22


















              1












              $begingroup$

              Hint $ $ Subtracting $, C = 5,$ then cancelling $x$ yields



              $qquad f(x) := x^{large n-1} = B + Ax + (x+5)^{large 2} p(x) $



              Thus $, B + Ax = f(-5) + f'(-5)(x+5),$ by Taylor expansion at $,x = -5,$






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                If you don't know (or can't use) Taylor then you could instead apply this (purely algebraic) Polynomial Double Root Test. on $,x^{n-1}!-B-Ax,,$ or just differentiate as I hinted on the question.
                $endgroup$
                – Bill Dubuque
                Jan 29 at 20:22
















              1












              1








              1





              $begingroup$

              Hint $ $ Subtracting $, C = 5,$ then cancelling $x$ yields



              $qquad f(x) := x^{large n-1} = B + Ax + (x+5)^{large 2} p(x) $



              Thus $, B + Ax = f(-5) + f'(-5)(x+5),$ by Taylor expansion at $,x = -5,$






              share|cite|improve this answer









              $endgroup$



              Hint $ $ Subtracting $, C = 5,$ then cancelling $x$ yields



              $qquad f(x) := x^{large n-1} = B + Ax + (x+5)^{large 2} p(x) $



              Thus $, B + Ax = f(-5) + f'(-5)(x+5),$ by Taylor expansion at $,x = -5,$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 29 at 20:12









              Bill DubuqueBill Dubuque

              213k29196654




              213k29196654












              • $begingroup$
                If you don't know (or can't use) Taylor then you could instead apply this (purely algebraic) Polynomial Double Root Test. on $,x^{n-1}!-B-Ax,,$ or just differentiate as I hinted on the question.
                $endgroup$
                – Bill Dubuque
                Jan 29 at 20:22




















              • $begingroup$
                If you don't know (or can't use) Taylor then you could instead apply this (purely algebraic) Polynomial Double Root Test. on $,x^{n-1}!-B-Ax,,$ or just differentiate as I hinted on the question.
                $endgroup$
                – Bill Dubuque
                Jan 29 at 20:22


















              $begingroup$
              If you don't know (or can't use) Taylor then you could instead apply this (purely algebraic) Polynomial Double Root Test. on $,x^{n-1}!-B-Ax,,$ or just differentiate as I hinted on the question.
              $endgroup$
              – Bill Dubuque
              Jan 29 at 20:22






              $begingroup$
              If you don't know (or can't use) Taylor then you could instead apply this (purely algebraic) Polynomial Double Root Test. on $,x^{n-1}!-B-Ax,,$ or just differentiate as I hinted on the question.
              $endgroup$
              – Bill Dubuque
              Jan 29 at 20:22













              0












              $begingroup$

              I think there is an error in the attempt that is part of the question. Anyway, here are some relevant quotients and remainders:



              $$ left( x^{4} + 5 right) = left( x^{3} + 10 x^{2} + 25 x right) cdot left( x - 10 right) + left( 75 x^{2} + 250 x + 5 right) $$
              $$ left( x^{5} + 5 right) = left( x^{3} + 10 x^{2} + 25 x right) cdot left( x^{2} - 10 x + 75 right) + left( - 500 x^{2} - 1875 x + 5 right) $$
              $$ left( x^{6} + 5 right) = left( x^{3} + 10 x^{2} + 25 x right) cdot left( x^{3} - 10 x^{2} + 75 x - 500 right) + left( 3125 x^{2} + 12500 x + 5 right) $$
              $$ left( x^{7} + 5 right) = left( x^{3} + 10 x^{2} + 25 x right) cdot left( x^{4} - 10 x^{3} + 75 x^{2} - 500 x + 3125 right) + left( - 18750 x^{2} - 78125 x + 5 right) $$
              $$ left( x^{8} + 5 right) = left( x^{3} + 10 x^{2} + 25 x right) cdot left( x^{5} - 10 x^{4} + 75 x^{3} - 500 x^{2} + 3125 x - 18750 right) + left( 109375 x^{2} + 468750 x + 5 right) $$
              $$ left( x^{9} + 5 right) = left( x^{3} + 10 x^{2} + 25 x right) cdot left( x^{6} - 10 x^{5} + 75 x^{4} - 500 x^{3} + 3125 x^{2} - 18750 x + 109375 right) + left( - 625000 x^{2} - 2734375 x + 5 right) $$
              $$ left( x^{10} + 5 right) = left( x^{3} + 10 x^{2} + 25 x right) cdot left( x^{7} - 10 x^{6} + 75 x^{5} - 500 x^{4} + 3125 x^{3} - 18750 x^{2} + 109375 x - 625000 right) + left( 3515625 x^{2} + 15625000 x + 5 right) $$






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                I think there is an error in the attempt that is part of the question. Anyway, here are some relevant quotients and remainders:



                $$ left( x^{4} + 5 right) = left( x^{3} + 10 x^{2} + 25 x right) cdot left( x - 10 right) + left( 75 x^{2} + 250 x + 5 right) $$
                $$ left( x^{5} + 5 right) = left( x^{3} + 10 x^{2} + 25 x right) cdot left( x^{2} - 10 x + 75 right) + left( - 500 x^{2} - 1875 x + 5 right) $$
                $$ left( x^{6} + 5 right) = left( x^{3} + 10 x^{2} + 25 x right) cdot left( x^{3} - 10 x^{2} + 75 x - 500 right) + left( 3125 x^{2} + 12500 x + 5 right) $$
                $$ left( x^{7} + 5 right) = left( x^{3} + 10 x^{2} + 25 x right) cdot left( x^{4} - 10 x^{3} + 75 x^{2} - 500 x + 3125 right) + left( - 18750 x^{2} - 78125 x + 5 right) $$
                $$ left( x^{8} + 5 right) = left( x^{3} + 10 x^{2} + 25 x right) cdot left( x^{5} - 10 x^{4} + 75 x^{3} - 500 x^{2} + 3125 x - 18750 right) + left( 109375 x^{2} + 468750 x + 5 right) $$
                $$ left( x^{9} + 5 right) = left( x^{3} + 10 x^{2} + 25 x right) cdot left( x^{6} - 10 x^{5} + 75 x^{4} - 500 x^{3} + 3125 x^{2} - 18750 x + 109375 right) + left( - 625000 x^{2} - 2734375 x + 5 right) $$
                $$ left( x^{10} + 5 right) = left( x^{3} + 10 x^{2} + 25 x right) cdot left( x^{7} - 10 x^{6} + 75 x^{5} - 500 x^{4} + 3125 x^{3} - 18750 x^{2} + 109375 x - 625000 right) + left( 3515625 x^{2} + 15625000 x + 5 right) $$






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  I think there is an error in the attempt that is part of the question. Anyway, here are some relevant quotients and remainders:



                  $$ left( x^{4} + 5 right) = left( x^{3} + 10 x^{2} + 25 x right) cdot left( x - 10 right) + left( 75 x^{2} + 250 x + 5 right) $$
                  $$ left( x^{5} + 5 right) = left( x^{3} + 10 x^{2} + 25 x right) cdot left( x^{2} - 10 x + 75 right) + left( - 500 x^{2} - 1875 x + 5 right) $$
                  $$ left( x^{6} + 5 right) = left( x^{3} + 10 x^{2} + 25 x right) cdot left( x^{3} - 10 x^{2} + 75 x - 500 right) + left( 3125 x^{2} + 12500 x + 5 right) $$
                  $$ left( x^{7} + 5 right) = left( x^{3} + 10 x^{2} + 25 x right) cdot left( x^{4} - 10 x^{3} + 75 x^{2} - 500 x + 3125 right) + left( - 18750 x^{2} - 78125 x + 5 right) $$
                  $$ left( x^{8} + 5 right) = left( x^{3} + 10 x^{2} + 25 x right) cdot left( x^{5} - 10 x^{4} + 75 x^{3} - 500 x^{2} + 3125 x - 18750 right) + left( 109375 x^{2} + 468750 x + 5 right) $$
                  $$ left( x^{9} + 5 right) = left( x^{3} + 10 x^{2} + 25 x right) cdot left( x^{6} - 10 x^{5} + 75 x^{4} - 500 x^{3} + 3125 x^{2} - 18750 x + 109375 right) + left( - 625000 x^{2} - 2734375 x + 5 right) $$
                  $$ left( x^{10} + 5 right) = left( x^{3} + 10 x^{2} + 25 x right) cdot left( x^{7} - 10 x^{6} + 75 x^{5} - 500 x^{4} + 3125 x^{3} - 18750 x^{2} + 109375 x - 625000 right) + left( 3515625 x^{2} + 15625000 x + 5 right) $$






                  share|cite|improve this answer











                  $endgroup$



                  I think there is an error in the attempt that is part of the question. Anyway, here are some relevant quotients and remainders:



                  $$ left( x^{4} + 5 right) = left( x^{3} + 10 x^{2} + 25 x right) cdot left( x - 10 right) + left( 75 x^{2} + 250 x + 5 right) $$
                  $$ left( x^{5} + 5 right) = left( x^{3} + 10 x^{2} + 25 x right) cdot left( x^{2} - 10 x + 75 right) + left( - 500 x^{2} - 1875 x + 5 right) $$
                  $$ left( x^{6} + 5 right) = left( x^{3} + 10 x^{2} + 25 x right) cdot left( x^{3} - 10 x^{2} + 75 x - 500 right) + left( 3125 x^{2} + 12500 x + 5 right) $$
                  $$ left( x^{7} + 5 right) = left( x^{3} + 10 x^{2} + 25 x right) cdot left( x^{4} - 10 x^{3} + 75 x^{2} - 500 x + 3125 right) + left( - 18750 x^{2} - 78125 x + 5 right) $$
                  $$ left( x^{8} + 5 right) = left( x^{3} + 10 x^{2} + 25 x right) cdot left( x^{5} - 10 x^{4} + 75 x^{3} - 500 x^{2} + 3125 x - 18750 right) + left( 109375 x^{2} + 468750 x + 5 right) $$
                  $$ left( x^{9} + 5 right) = left( x^{3} + 10 x^{2} + 25 x right) cdot left( x^{6} - 10 x^{5} + 75 x^{4} - 500 x^{3} + 3125 x^{2} - 18750 x + 109375 right) + left( - 625000 x^{2} - 2734375 x + 5 right) $$
                  $$ left( x^{10} + 5 right) = left( x^{3} + 10 x^{2} + 25 x right) cdot left( x^{7} - 10 x^{6} + 75 x^{5} - 500 x^{4} + 3125 x^{3} - 18750 x^{2} + 109375 x - 625000 right) + left( 3515625 x^{2} + 15625000 x + 5 right) $$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 30 at 0:08

























                  answered Jan 29 at 21:51









                  Will JagyWill Jagy

                  104k5102201




                  104k5102201






























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