How to go from polynomial of third degree to multiplication of two smaller polynomials












1












$begingroup$


I have a basic calculus question which I should be able to do easily but I just can't remember how to tackle it. I'm working on my linear algebra exam and trying to find the eigenvalues for a certain matrix A. I got the following as the determinant:



$-lambda^{3}+3lambda^{2}+9lambda-27$ and I want to get this in the form of $(lambda+x_{1})^{2}(lambda+x_{2})$. How to get there?



PS: the solution for this particular one is $x_{1}=-3$ and $x_{2}=3$, it case that might help you.






algebra-precalculus polynomials







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  • 1




    $begingroup$
    The form giving the roots is $(lambda-x_{1})^{2}(lambda-x_{2}).$ In the present case are roots $3,-3$ so this will have no effect, but in general be careful.
    $endgroup$
    – user376343
    Dec 18 '18 at 13:00












  • $begingroup$
    Good point, thanks
    $endgroup$
    – Mathbeginner
    Dec 18 '18 at 14:25
















1












$begingroup$


I have a basic calculus question which I should be able to do easily but I just can't remember how to tackle it. I'm working on my linear algebra exam and trying to find the eigenvalues for a certain matrix A. I got the following as the determinant:



$-lambda^{3}+3lambda^{2}+9lambda-27$ and I want to get this in the form of $(lambda+x_{1})^{2}(lambda+x_{2})$. How to get there?



PS: the solution for this particular one is $x_{1}=-3$ and $x_{2}=3$, it case that might help you.






algebra-precalculus polynomials







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The form giving the roots is $(lambda-x_{1})^{2}(lambda-x_{2}).$ In the present case are roots $3,-3$ so this will have no effect, but in general be careful.
    $endgroup$
    – user376343
    Dec 18 '18 at 13:00












  • $begingroup$
    Good point, thanks
    $endgroup$
    – Mathbeginner
    Dec 18 '18 at 14:25














1












1








1





$begingroup$


I have a basic calculus question which I should be able to do easily but I just can't remember how to tackle it. I'm working on my linear algebra exam and trying to find the eigenvalues for a certain matrix A. I got the following as the determinant:



$-lambda^{3}+3lambda^{2}+9lambda-27$ and I want to get this in the form of $(lambda+x_{1})^{2}(lambda+x_{2})$. How to get there?



PS: the solution for this particular one is $x_{1}=-3$ and $x_{2}=3$, it case that might help you.






algebra-precalculus polynomials







share|cite|improve this question











$endgroup$




I have a basic calculus question which I should be able to do easily but I just can't remember how to tackle it. I'm working on my linear algebra exam and trying to find the eigenvalues for a certain matrix A. I got the following as the determinant:



$-lambda^{3}+3lambda^{2}+9lambda-27$ and I want to get this in the form of $(lambda+x_{1})^{2}(lambda+x_{2})$. How to get there?



PS: the solution for this particular one is $x_{1}=-3$ and $x_{2}=3$, it case that might help you.






algebra-precalculus polynomials




algebra-precalculus polynomials






share|cite|improve this question















share|cite|improve this question













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share|cite|improve this question








edited Dec 18 '18 at 12:57









Jyrki Lahtonen

110k13171385




110k13171385










asked Dec 18 '18 at 11:56









MathbeginnerMathbeginner

1908




1908








  • 1




    $begingroup$
    The form giving the roots is $(lambda-x_{1})^{2}(lambda-x_{2}).$ In the present case are roots $3,-3$ so this will have no effect, but in general be careful.
    $endgroup$
    – user376343
    Dec 18 '18 at 13:00












  • $begingroup$
    Good point, thanks
    $endgroup$
    – Mathbeginner
    Dec 18 '18 at 14:25














  • 1




    $begingroup$
    The form giving the roots is $(lambda-x_{1})^{2}(lambda-x_{2}).$ In the present case are roots $3,-3$ so this will have no effect, but in general be careful.
    $endgroup$
    – user376343
    Dec 18 '18 at 13:00












  • $begingroup$
    Good point, thanks
    $endgroup$
    – Mathbeginner
    Dec 18 '18 at 14:25








1




1




$begingroup$
The form giving the roots is $(lambda-x_{1})^{2}(lambda-x_{2}).$ In the present case are roots $3,-3$ so this will have no effect, but in general be careful.
$endgroup$
– user376343
Dec 18 '18 at 13:00






$begingroup$
The form giving the roots is $(lambda-x_{1})^{2}(lambda-x_{2}).$ In the present case are roots $3,-3$ so this will have no effect, but in general be careful.
$endgroup$
– user376343
Dec 18 '18 at 13:00














$begingroup$
Good point, thanks
$endgroup$
– Mathbeginner
Dec 18 '18 at 14:25




$begingroup$
Good point, thanks
$endgroup$
– Mathbeginner
Dec 18 '18 at 14:25










2 Answers
2






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$begingroup$

Using the rational root theorem, you can find those roots that you mentioned. They are $pm3$, as you know. So, divide your original polynomial by $lambda-3$ and then divide what you got by $lambda+3$.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    The polynomial can be factored by grouping:
    $$
    -lambda^{3}+3lambda^{2}+9lambda-27
    = -lambda^{2}(lambda-3)+9(lambda-3)
    = -(lambda^{2}-9)(lambda-3)
    = -(lambda+3)(lambda-3)^{2}
    $$






    share|cite|improve this answer











    $endgroup$













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      2 Answers
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      2 Answers
      2






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      1












      $begingroup$

      Using the rational root theorem, you can find those roots that you mentioned. They are $pm3$, as you know. So, divide your original polynomial by $lambda-3$ and then divide what you got by $lambda+3$.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Using the rational root theorem, you can find those roots that you mentioned. They are $pm3$, as you know. So, divide your original polynomial by $lambda-3$ and then divide what you got by $lambda+3$.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Using the rational root theorem, you can find those roots that you mentioned. They are $pm3$, as you know. So, divide your original polynomial by $lambda-3$ and then divide what you got by $lambda+3$.






          share|cite|improve this answer









          $endgroup$



          Using the rational root theorem, you can find those roots that you mentioned. They are $pm3$, as you know. So, divide your original polynomial by $lambda-3$ and then divide what you got by $lambda+3$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 18 '18 at 12:00









          José Carlos SantosJosé Carlos Santos

          168k22132236




          168k22132236























              1












              $begingroup$

              The polynomial can be factored by grouping:
              $$
              -lambda^{3}+3lambda^{2}+9lambda-27
              = -lambda^{2}(lambda-3)+9(lambda-3)
              = -(lambda^{2}-9)(lambda-3)
              = -(lambda+3)(lambda-3)^{2}
              $$






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                The polynomial can be factored by grouping:
                $$
                -lambda^{3}+3lambda^{2}+9lambda-27
                = -lambda^{2}(lambda-3)+9(lambda-3)
                = -(lambda^{2}-9)(lambda-3)
                = -(lambda+3)(lambda-3)^{2}
                $$






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  The polynomial can be factored by grouping:
                  $$
                  -lambda^{3}+3lambda^{2}+9lambda-27
                  = -lambda^{2}(lambda-3)+9(lambda-3)
                  = -(lambda^{2}-9)(lambda-3)
                  = -(lambda+3)(lambda-3)^{2}
                  $$






                  share|cite|improve this answer











                  $endgroup$



                  The polynomial can be factored by grouping:
                  $$
                  -lambda^{3}+3lambda^{2}+9lambda-27
                  = -lambda^{2}(lambda-3)+9(lambda-3)
                  = -(lambda^{2}-9)(lambda-3)
                  = -(lambda+3)(lambda-3)^{2}
                  $$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 19 '18 at 11:03

























                  answered Dec 18 '18 at 14:06









                  lhflhf

                  166k10171400




                  166k10171400






























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