How do I factor this complex polynomial?












0












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My math teacher offered my class extra credit if we could factor this equation. It is not factorable by grouping. If you put the values into a polynomial root solver the zeros are Integer.



Equation: $x(x^3 + 4x^2 -3x -18) = 0$






algebra-precalculus







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  • 7




    $begingroup$
    Are you familiar with the rational root theorem?
    $endgroup$
    – platty
    Dec 13 '18 at 20:44










  • $begingroup$
    there were both + - with the term 3x. What has to be there?
    $endgroup$
    – user376343
    Dec 13 '18 at 20:45










  • $begingroup$
    Why don’t you try to find small integers that are roots of the polynomial?
    $endgroup$
    – Mindlack
    Dec 13 '18 at 20:48










  • $begingroup$
    Because you have $x$ as a factor, there is one trivial root: $x(x^3+4x^2-3x-18)=0implies x=0 or x^3+4x^2-3x-18=0$ The remaining roots are the solutions to the polynomial $x^3+4x^2-3x-18=0$
    $endgroup$
    – R. Burton
    Dec 13 '18 at 22:44
















0












$begingroup$


My math teacher offered my class extra credit if we could factor this equation. It is not factorable by grouping. If you put the values into a polynomial root solver the zeros are Integer.



Equation: $x(x^3 + 4x^2 -3x -18) = 0$






algebra-precalculus







share|cite|improve this question











$endgroup$








  • 7




    $begingroup$
    Are you familiar with the rational root theorem?
    $endgroup$
    – platty
    Dec 13 '18 at 20:44










  • $begingroup$
    there were both + - with the term 3x. What has to be there?
    $endgroup$
    – user376343
    Dec 13 '18 at 20:45










  • $begingroup$
    Why don’t you try to find small integers that are roots of the polynomial?
    $endgroup$
    – Mindlack
    Dec 13 '18 at 20:48










  • $begingroup$
    Because you have $x$ as a factor, there is one trivial root: $x(x^3+4x^2-3x-18)=0implies x=0 or x^3+4x^2-3x-18=0$ The remaining roots are the solutions to the polynomial $x^3+4x^2-3x-18=0$
    $endgroup$
    – R. Burton
    Dec 13 '18 at 22:44














0












0








0





$begingroup$


My math teacher offered my class extra credit if we could factor this equation. It is not factorable by grouping. If you put the values into a polynomial root solver the zeros are Integer.



Equation: $x(x^3 + 4x^2 -3x -18) = 0$






algebra-precalculus







share|cite|improve this question











$endgroup$




My math teacher offered my class extra credit if we could factor this equation. It is not factorable by grouping. If you put the values into a polynomial root solver the zeros are Integer.



Equation: $x(x^3 + 4x^2 -3x -18) = 0$






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 Dec 13 '18 at 20:44









user376343

3,7883828




3,7883828










asked Dec 13 '18 at 20:43









xXNebulaNinjaXxxXNebulaNinjaXx

31




31








  • 7




    $begingroup$
    Are you familiar with the rational root theorem?
    $endgroup$
    – platty
    Dec 13 '18 at 20:44










  • $begingroup$
    there were both + - with the term 3x. What has to be there?
    $endgroup$
    – user376343
    Dec 13 '18 at 20:45










  • $begingroup$
    Why don’t you try to find small integers that are roots of the polynomial?
    $endgroup$
    – Mindlack
    Dec 13 '18 at 20:48










  • $begingroup$
    Because you have $x$ as a factor, there is one trivial root: $x(x^3+4x^2-3x-18)=0implies x=0 or x^3+4x^2-3x-18=0$ The remaining roots are the solutions to the polynomial $x^3+4x^2-3x-18=0$
    $endgroup$
    – R. Burton
    Dec 13 '18 at 22:44














  • 7




    $begingroup$
    Are you familiar with the rational root theorem?
    $endgroup$
    – platty
    Dec 13 '18 at 20:44










  • $begingroup$
    there were both + - with the term 3x. What has to be there?
    $endgroup$
    – user376343
    Dec 13 '18 at 20:45










  • $begingroup$
    Why don’t you try to find small integers that are roots of the polynomial?
    $endgroup$
    – Mindlack
    Dec 13 '18 at 20:48










  • $begingroup$
    Because you have $x$ as a factor, there is one trivial root: $x(x^3+4x^2-3x-18)=0implies x=0 or x^3+4x^2-3x-18=0$ The remaining roots are the solutions to the polynomial $x^3+4x^2-3x-18=0$
    $endgroup$
    – R. Burton
    Dec 13 '18 at 22:44








7




7




$begingroup$
Are you familiar with the rational root theorem?
$endgroup$
– platty
Dec 13 '18 at 20:44




$begingroup$
Are you familiar with the rational root theorem?
$endgroup$
– platty
Dec 13 '18 at 20:44












$begingroup$
there were both + - with the term 3x. What has to be there?
$endgroup$
– user376343
Dec 13 '18 at 20:45




$begingroup$
there were both + - with the term 3x. What has to be there?
$endgroup$
– user376343
Dec 13 '18 at 20:45












$begingroup$
Why don’t you try to find small integers that are roots of the polynomial?
$endgroup$
– Mindlack
Dec 13 '18 at 20:48




$begingroup$
Why don’t you try to find small integers that are roots of the polynomial?
$endgroup$
– Mindlack
Dec 13 '18 at 20:48












$begingroup$
Because you have $x$ as a factor, there is one trivial root: $x(x^3+4x^2-3x-18)=0implies x=0 or x^3+4x^2-3x-18=0$ The remaining roots are the solutions to the polynomial $x^3+4x^2-3x-18=0$
$endgroup$
– R. Burton
Dec 13 '18 at 22:44




$begingroup$
Because you have $x$ as a factor, there is one trivial root: $x(x^3+4x^2-3x-18)=0implies x=0 or x^3+4x^2-3x-18=0$ The remaining roots are the solutions to the polynomial $x^3+4x^2-3x-18=0$
$endgroup$
– R. Burton
Dec 13 '18 at 22:44










2 Answers
2






active

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1












$begingroup$

Adding to R. Burton's comment, if you solve $x^3+4x^2-3x-18$,by Rational Root Theorem, possible roots are $pm1,pm2,pm3,pm6,pm9,pm18$. Observe that positive $2$ is a root. Therefore, $x^3+4x^2-3x-18=(x-2)q(x)$ where $q(x)$ is quadratic. Now, $q(x)=(x+3)^2$, giving you all other three roots.
Hence, $x(x^3+4x^2-3x-18)=x(x-2){(x+3)}^2$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Corrected: $q(x)=(x+3)^2$
    $endgroup$
    – z100
    Dec 14 '18 at 14:04










  • $begingroup$
    @z100 Sorry, my bad.
    $endgroup$
    – AryanSonwatikar
    Dec 14 '18 at 15:21



















0












$begingroup$

$x^3+4x^2-3x-18 = (x^3 +3x^2)+(x^2-3x-18) = x^2(x+3)+(x+3)(x-6)=(x+3)(x^2+x-6)=(x+3)(x+3)(x-2)$






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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

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    1












    $begingroup$

    Adding to R. Burton's comment, if you solve $x^3+4x^2-3x-18$,by Rational Root Theorem, possible roots are $pm1,pm2,pm3,pm6,pm9,pm18$. Observe that positive $2$ is a root. Therefore, $x^3+4x^2-3x-18=(x-2)q(x)$ where $q(x)$ is quadratic. Now, $q(x)=(x+3)^2$, giving you all other three roots.
    Hence, $x(x^3+4x^2-3x-18)=x(x-2){(x+3)}^2$






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Corrected: $q(x)=(x+3)^2$
      $endgroup$
      – z100
      Dec 14 '18 at 14:04










    • $begingroup$
      @z100 Sorry, my bad.
      $endgroup$
      – AryanSonwatikar
      Dec 14 '18 at 15:21
















    1












    $begingroup$

    Adding to R. Burton's comment, if you solve $x^3+4x^2-3x-18$,by Rational Root Theorem, possible roots are $pm1,pm2,pm3,pm6,pm9,pm18$. Observe that positive $2$ is a root. Therefore, $x^3+4x^2-3x-18=(x-2)q(x)$ where $q(x)$ is quadratic. Now, $q(x)=(x+3)^2$, giving you all other three roots.
    Hence, $x(x^3+4x^2-3x-18)=x(x-2){(x+3)}^2$






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Corrected: $q(x)=(x+3)^2$
      $endgroup$
      – z100
      Dec 14 '18 at 14:04










    • $begingroup$
      @z100 Sorry, my bad.
      $endgroup$
      – AryanSonwatikar
      Dec 14 '18 at 15:21














    1












    1








    1





    $begingroup$

    Adding to R. Burton's comment, if you solve $x^3+4x^2-3x-18$,by Rational Root Theorem, possible roots are $pm1,pm2,pm3,pm6,pm9,pm18$. Observe that positive $2$ is a root. Therefore, $x^3+4x^2-3x-18=(x-2)q(x)$ where $q(x)$ is quadratic. Now, $q(x)=(x+3)^2$, giving you all other three roots.
    Hence, $x(x^3+4x^2-3x-18)=x(x-2){(x+3)}^2$






    share|cite|improve this answer











    $endgroup$



    Adding to R. Burton's comment, if you solve $x^3+4x^2-3x-18$,by Rational Root Theorem, possible roots are $pm1,pm2,pm3,pm6,pm9,pm18$. Observe that positive $2$ is a root. Therefore, $x^3+4x^2-3x-18=(x-2)q(x)$ where $q(x)$ is quadratic. Now, $q(x)=(x+3)^2$, giving you all other three roots.
    Hence, $x(x^3+4x^2-3x-18)=x(x-2){(x+3)}^2$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 14 '18 at 15:20

























    answered Dec 14 '18 at 12:14









    AryanSonwatikarAryanSonwatikar

    453114




    453114








    • 1




      $begingroup$
      Corrected: $q(x)=(x+3)^2$
      $endgroup$
      – z100
      Dec 14 '18 at 14:04










    • $begingroup$
      @z100 Sorry, my bad.
      $endgroup$
      – AryanSonwatikar
      Dec 14 '18 at 15:21














    • 1




      $begingroup$
      Corrected: $q(x)=(x+3)^2$
      $endgroup$
      – z100
      Dec 14 '18 at 14:04










    • $begingroup$
      @z100 Sorry, my bad.
      $endgroup$
      – AryanSonwatikar
      Dec 14 '18 at 15:21








    1




    1




    $begingroup$
    Corrected: $q(x)=(x+3)^2$
    $endgroup$
    – z100
    Dec 14 '18 at 14:04




    $begingroup$
    Corrected: $q(x)=(x+3)^2$
    $endgroup$
    – z100
    Dec 14 '18 at 14:04












    $begingroup$
    @z100 Sorry, my bad.
    $endgroup$
    – AryanSonwatikar
    Dec 14 '18 at 15:21




    $begingroup$
    @z100 Sorry, my bad.
    $endgroup$
    – AryanSonwatikar
    Dec 14 '18 at 15:21











    0












    $begingroup$

    $x^3+4x^2-3x-18 = (x^3 +3x^2)+(x^2-3x-18) = x^2(x+3)+(x+3)(x-6)=(x+3)(x^2+x-6)=(x+3)(x+3)(x-2)$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      $x^3+4x^2-3x-18 = (x^3 +3x^2)+(x^2-3x-18) = x^2(x+3)+(x+3)(x-6)=(x+3)(x^2+x-6)=(x+3)(x+3)(x-2)$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        $x^3+4x^2-3x-18 = (x^3 +3x^2)+(x^2-3x-18) = x^2(x+3)+(x+3)(x-6)=(x+3)(x^2+x-6)=(x+3)(x+3)(x-2)$






        share|cite|improve this answer









        $endgroup$



        $x^3+4x^2-3x-18 = (x^3 +3x^2)+(x^2-3x-18) = x^2(x+3)+(x+3)(x-6)=(x+3)(x^2+x-6)=(x+3)(x+3)(x-2)$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 14 '18 at 13:03









        z100z100

        48848




        48848






























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