How do you factor: $2x^4+x^3-7x^2-x+5$ [closed]












0












$begingroup$


I don’t know how factor this, I’ve never factored something with a degree higher than 3.






factoring







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$endgroup$



closed as off-topic by amWhy, anomaly, Shailesh, José Carlos Santos, KReiser Jan 11 at 9:05


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, anomaly, Shailesh, José Carlos Santos, KReiser

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 2




    $begingroup$
    I would start by trying rational roots. Do you know the rational root theorem, and what that would imply about any rational roots (if they exist)?
    $endgroup$
    – Doug M
    Jan 11 at 0:19












  • $begingroup$
    Look at the coefficients and add them up....you have $$2x^4+x^3-7x^2-x+5$$ So $2+1-7-1+5=...$ what does that mean?
    $endgroup$
    – Lalaloopsy
    Jan 11 at 0:27












  • $begingroup$
    Hint: $f (1)=0$. So $x-1$ factors out. And then you are left with a polynomial of degree $3$. Which maybe you can factor?
    $endgroup$
    – fleablood
    Jan 11 at 0:27






  • 1




    $begingroup$
    $(x-1) (x+1) left(2 x^2+x-5right)$
    $endgroup$
    – David G. Stork
    Jan 11 at 0:30
















0












$begingroup$


I don’t know how factor this, I’ve never factored something with a degree higher than 3.






factoring







share|cite|improve this question











$endgroup$



closed as off-topic by amWhy, anomaly, Shailesh, José Carlos Santos, KReiser Jan 11 at 9:05


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, anomaly, Shailesh, José Carlos Santos, KReiser

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 2




    $begingroup$
    I would start by trying rational roots. Do you know the rational root theorem, and what that would imply about any rational roots (if they exist)?
    $endgroup$
    – Doug M
    Jan 11 at 0:19












  • $begingroup$
    Look at the coefficients and add them up....you have $$2x^4+x^3-7x^2-x+5$$ So $2+1-7-1+5=...$ what does that mean?
    $endgroup$
    – Lalaloopsy
    Jan 11 at 0:27












  • $begingroup$
    Hint: $f (1)=0$. So $x-1$ factors out. And then you are left with a polynomial of degree $3$. Which maybe you can factor?
    $endgroup$
    – fleablood
    Jan 11 at 0:27






  • 1




    $begingroup$
    $(x-1) (x+1) left(2 x^2+x-5right)$
    $endgroup$
    – David G. Stork
    Jan 11 at 0:30














0












0








0





$begingroup$


I don’t know how factor this, I’ve never factored something with a degree higher than 3.






factoring







share|cite|improve this question











$endgroup$




I don’t know how factor this, I’ve never factored something with a degree higher than 3.






factoring




factoring






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 11 at 0:30









David G. Stork

10.7k31332




10.7k31332










asked Jan 11 at 0:16









annaîs mendoxaannaîs mendoxa

1




1




closed as off-topic by amWhy, anomaly, Shailesh, José Carlos Santos, KReiser Jan 11 at 9:05


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, anomaly, Shailesh, José Carlos Santos, KReiser

If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by amWhy, anomaly, Shailesh, José Carlos Santos, KReiser Jan 11 at 9:05


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, anomaly, Shailesh, José Carlos Santos, KReiser

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 2




    $begingroup$
    I would start by trying rational roots. Do you know the rational root theorem, and what that would imply about any rational roots (if they exist)?
    $endgroup$
    – Doug M
    Jan 11 at 0:19












  • $begingroup$
    Look at the coefficients and add them up....you have $$2x^4+x^3-7x^2-x+5$$ So $2+1-7-1+5=...$ what does that mean?
    $endgroup$
    – Lalaloopsy
    Jan 11 at 0:27












  • $begingroup$
    Hint: $f (1)=0$. So $x-1$ factors out. And then you are left with a polynomial of degree $3$. Which maybe you can factor?
    $endgroup$
    – fleablood
    Jan 11 at 0:27






  • 1




    $begingroup$
    $(x-1) (x+1) left(2 x^2+x-5right)$
    $endgroup$
    – David G. Stork
    Jan 11 at 0:30














  • 2




    $begingroup$
    I would start by trying rational roots. Do you know the rational root theorem, and what that would imply about any rational roots (if they exist)?
    $endgroup$
    – Doug M
    Jan 11 at 0:19












  • $begingroup$
    Look at the coefficients and add them up....you have $$2x^4+x^3-7x^2-x+5$$ So $2+1-7-1+5=...$ what does that mean?
    $endgroup$
    – Lalaloopsy
    Jan 11 at 0:27












  • $begingroup$
    Hint: $f (1)=0$. So $x-1$ factors out. And then you are left with a polynomial of degree $3$. Which maybe you can factor?
    $endgroup$
    – fleablood
    Jan 11 at 0:27






  • 1




    $begingroup$
    $(x-1) (x+1) left(2 x^2+x-5right)$
    $endgroup$
    – David G. Stork
    Jan 11 at 0:30








2




2




$begingroup$
I would start by trying rational roots. Do you know the rational root theorem, and what that would imply about any rational roots (if they exist)?
$endgroup$
– Doug M
Jan 11 at 0:19






$begingroup$
I would start by trying rational roots. Do you know the rational root theorem, and what that would imply about any rational roots (if they exist)?
$endgroup$
– Doug M
Jan 11 at 0:19














$begingroup$
Look at the coefficients and add them up....you have $$2x^4+x^3-7x^2-x+5$$ So $2+1-7-1+5=...$ what does that mean?
$endgroup$
– Lalaloopsy
Jan 11 at 0:27






$begingroup$
Look at the coefficients and add them up....you have $$2x^4+x^3-7x^2-x+5$$ So $2+1-7-1+5=...$ what does that mean?
$endgroup$
– Lalaloopsy
Jan 11 at 0:27














$begingroup$
Hint: $f (1)=0$. So $x-1$ factors out. And then you are left with a polynomial of degree $3$. Which maybe you can factor?
$endgroup$
– fleablood
Jan 11 at 0:27




$begingroup$
Hint: $f (1)=0$. So $x-1$ factors out. And then you are left with a polynomial of degree $3$. Which maybe you can factor?
$endgroup$
– fleablood
Jan 11 at 0:27




1




1




$begingroup$
$(x-1) (x+1) left(2 x^2+x-5right)$
$endgroup$
– David G. Stork
Jan 11 at 0:30




$begingroup$
$(x-1) (x+1) left(2 x^2+x-5right)$
$endgroup$
– David G. Stork
Jan 11 at 0:30










4 Answers
4






active

oldest

votes


















1












$begingroup$

In this case, we can start by applying the rational root theorem to determine possible rational roots. In this case there are 4 total roots.



You can study the rational root theorem here https://en.wikipedia.org/wiki/Rational_root_theorem.



The possible root combinations in this case are $-1,1,-frac{1}{2},frac{1}{2},-5,5,-frac{5}{2}$, and $frac{5}{2}$.



A look at this function's graph reveals that $x=-1$ is a root with a multiplicity of 1. The graph also reveals that the function's 4 roots are real. This can also be confirmed analytically by Descarte's Rule of signs.



Since we know that $x=-1$ is a root, we can divide the function synthetically by $-1$ to get a cubic function. From there, you can either factor the cubic or perform synthetic division on the cubic with another factor to reduce it to a quadratic.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Hint $ $ It is $ (2x^4-7x^2+5) + (x^3-x), $ and both clearly have roots $,x = pm 1$






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      Hint:



      This polynoial is divisible by $x^2-1$:
      begin{align}
      2x^4+x^3-7x^2-x+5&=(2x^4-2x^2)+x^3-5x^2-x+5\
      &=2x^2(x^2-1)+ x^2(x-5)-x+5 \
      &= (x^2-1)(2x^2+x-5).
      end{align}      +






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        But you pulled the factor out of a hat.
        $endgroup$
        – Bill Dubuque
        Jan 11 at 0:35










      • $begingroup$
        The rational roots theorem helped for the possible quadratic factors.
        $endgroup$
        – Bernard
        Jan 11 at 0:36



















      0












      $begingroup$

      Starting with $2 x^4+x^3-7 x^2-x+5$ and using the insight that $x=1$ must be a factor:



      Divide out:



      $${2 x^4+x^3-7 x^2-x+5 over x-1} = 2 x^3+3 x^2-4 x-5$$



      Notice again, from the sum of coefficients that $x = -1$ is a root, so divide out again:



      $${2 x^3+3 x^2-4 x-5 over x+1} = 2 x^2+x-5.$$



      So:



      $$(x-1)(x+1)(2 x^2+x-5)$$



      Use the quadratic equation:



      $$(x-1)(x+1)(x - 1/4 (-1 - sqrt{41}))(x+1/4 (-1 - sqrt{41}))$$






      share|cite|improve this answer









      $endgroup$




















        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        1












        $begingroup$

        In this case, we can start by applying the rational root theorem to determine possible rational roots. In this case there are 4 total roots.



        You can study the rational root theorem here https://en.wikipedia.org/wiki/Rational_root_theorem.



        The possible root combinations in this case are $-1,1,-frac{1}{2},frac{1}{2},-5,5,-frac{5}{2}$, and $frac{5}{2}$.



        A look at this function's graph reveals that $x=-1$ is a root with a multiplicity of 1. The graph also reveals that the function's 4 roots are real. This can also be confirmed analytically by Descarte's Rule of signs.



        Since we know that $x=-1$ is a root, we can divide the function synthetically by $-1$ to get a cubic function. From there, you can either factor the cubic or perform synthetic division on the cubic with another factor to reduce it to a quadratic.






        share|cite|improve this answer









        $endgroup$


















          1












          $begingroup$

          In this case, we can start by applying the rational root theorem to determine possible rational roots. In this case there are 4 total roots.



          You can study the rational root theorem here https://en.wikipedia.org/wiki/Rational_root_theorem.



          The possible root combinations in this case are $-1,1,-frac{1}{2},frac{1}{2},-5,5,-frac{5}{2}$, and $frac{5}{2}$.



          A look at this function's graph reveals that $x=-1$ is a root with a multiplicity of 1. The graph also reveals that the function's 4 roots are real. This can also be confirmed analytically by Descarte's Rule of signs.



          Since we know that $x=-1$ is a root, we can divide the function synthetically by $-1$ to get a cubic function. From there, you can either factor the cubic or perform synthetic division on the cubic with another factor to reduce it to a quadratic.






          share|cite|improve this answer









          $endgroup$
















            1












            1








            1





            $begingroup$

            In this case, we can start by applying the rational root theorem to determine possible rational roots. In this case there are 4 total roots.



            You can study the rational root theorem here https://en.wikipedia.org/wiki/Rational_root_theorem.



            The possible root combinations in this case are $-1,1,-frac{1}{2},frac{1}{2},-5,5,-frac{5}{2}$, and $frac{5}{2}$.



            A look at this function's graph reveals that $x=-1$ is a root with a multiplicity of 1. The graph also reveals that the function's 4 roots are real. This can also be confirmed analytically by Descarte's Rule of signs.



            Since we know that $x=-1$ is a root, we can divide the function synthetically by $-1$ to get a cubic function. From there, you can either factor the cubic or perform synthetic division on the cubic with another factor to reduce it to a quadratic.






            share|cite|improve this answer









            $endgroup$



            In this case, we can start by applying the rational root theorem to determine possible rational roots. In this case there are 4 total roots.



            You can study the rational root theorem here https://en.wikipedia.org/wiki/Rational_root_theorem.



            The possible root combinations in this case are $-1,1,-frac{1}{2},frac{1}{2},-5,5,-frac{5}{2}$, and $frac{5}{2}$.



            A look at this function's graph reveals that $x=-1$ is a root with a multiplicity of 1. The graph also reveals that the function's 4 roots are real. This can also be confirmed analytically by Descarte's Rule of signs.



            Since we know that $x=-1$ is a root, we can divide the function synthetically by $-1$ to get a cubic function. From there, you can either factor the cubic or perform synthetic division on the cubic with another factor to reduce it to a quadratic.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 11 at 0:27









            GnumbertesterGnumbertester

            37518




            37518























                0












                $begingroup$

                Hint $ $ It is $ (2x^4-7x^2+5) + (x^3-x), $ and both clearly have roots $,x = pm 1$






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  Hint $ $ It is $ (2x^4-7x^2+5) + (x^3-x), $ and both clearly have roots $,x = pm 1$






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    Hint $ $ It is $ (2x^4-7x^2+5) + (x^3-x), $ and both clearly have roots $,x = pm 1$






                    share|cite|improve this answer









                    $endgroup$



                    Hint $ $ It is $ (2x^4-7x^2+5) + (x^3-x), $ and both clearly have roots $,x = pm 1$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 11 at 0:32









                    Bill DubuqueBill Dubuque

                    209k29191638




                    209k29191638























                        0












                        $begingroup$

                        Hint:



                        This polynoial is divisible by $x^2-1$:
                        begin{align}
                        2x^4+x^3-7x^2-x+5&=(2x^4-2x^2)+x^3-5x^2-x+5\
                        &=2x^2(x^2-1)+ x^2(x-5)-x+5 \
                        &= (x^2-1)(2x^2+x-5).
                        end{align}                        +






                        share|cite|improve this answer









                        $endgroup$













                        • $begingroup$
                          But you pulled the factor out of a hat.
                          $endgroup$
                          – Bill Dubuque
                          Jan 11 at 0:35










                        • $begingroup$
                          The rational roots theorem helped for the possible quadratic factors.
                          $endgroup$
                          – Bernard
                          Jan 11 at 0:36
















                        0












                        $begingroup$

                        Hint:



                        This polynoial is divisible by $x^2-1$:
                        begin{align}
                        2x^4+x^3-7x^2-x+5&=(2x^4-2x^2)+x^3-5x^2-x+5\
                        &=2x^2(x^2-1)+ x^2(x-5)-x+5 \
                        &= (x^2-1)(2x^2+x-5).
                        end{align}                        +






                        share|cite|improve this answer









                        $endgroup$













                        • $begingroup$
                          But you pulled the factor out of a hat.
                          $endgroup$
                          – Bill Dubuque
                          Jan 11 at 0:35










                        • $begingroup$
                          The rational roots theorem helped for the possible quadratic factors.
                          $endgroup$
                          – Bernard
                          Jan 11 at 0:36














                        0












                        0








                        0





                        $begingroup$

                        Hint:



                        This polynoial is divisible by $x^2-1$:
                        begin{align}
                        2x^4+x^3-7x^2-x+5&=(2x^4-2x^2)+x^3-5x^2-x+5\
                        &=2x^2(x^2-1)+ x^2(x-5)-x+5 \
                        &= (x^2-1)(2x^2+x-5).
                        end{align}                        +






                        share|cite|improve this answer









                        $endgroup$



                        Hint:



                        This polynoial is divisible by $x^2-1$:
                        begin{align}
                        2x^4+x^3-7x^2-x+5&=(2x^4-2x^2)+x^3-5x^2-x+5\
                        &=2x^2(x^2-1)+ x^2(x-5)-x+5 \
                        &= (x^2-1)(2x^2+x-5).
                        end{align}                        +







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered Jan 11 at 0:34









                        BernardBernard

                        119k740113




                        119k740113












                        • $begingroup$
                          But you pulled the factor out of a hat.
                          $endgroup$
                          – Bill Dubuque
                          Jan 11 at 0:35










                        • $begingroup$
                          The rational roots theorem helped for the possible quadratic factors.
                          $endgroup$
                          – Bernard
                          Jan 11 at 0:36


















                        • $begingroup$
                          But you pulled the factor out of a hat.
                          $endgroup$
                          – Bill Dubuque
                          Jan 11 at 0:35










                        • $begingroup$
                          The rational roots theorem helped for the possible quadratic factors.
                          $endgroup$
                          – Bernard
                          Jan 11 at 0:36
















                        $begingroup$
                        But you pulled the factor out of a hat.
                        $endgroup$
                        – Bill Dubuque
                        Jan 11 at 0:35




                        $begingroup$
                        But you pulled the factor out of a hat.
                        $endgroup$
                        – Bill Dubuque
                        Jan 11 at 0:35












                        $begingroup$
                        The rational roots theorem helped for the possible quadratic factors.
                        $endgroup$
                        – Bernard
                        Jan 11 at 0:36




                        $begingroup$
                        The rational roots theorem helped for the possible quadratic factors.
                        $endgroup$
                        – Bernard
                        Jan 11 at 0:36











                        0












                        $begingroup$

                        Starting with $2 x^4+x^3-7 x^2-x+5$ and using the insight that $x=1$ must be a factor:



                        Divide out:



                        $${2 x^4+x^3-7 x^2-x+5 over x-1} = 2 x^3+3 x^2-4 x-5$$



                        Notice again, from the sum of coefficients that $x = -1$ is a root, so divide out again:



                        $${2 x^3+3 x^2-4 x-5 over x+1} = 2 x^2+x-5.$$



                        So:



                        $$(x-1)(x+1)(2 x^2+x-5)$$



                        Use the quadratic equation:



                        $$(x-1)(x+1)(x - 1/4 (-1 - sqrt{41}))(x+1/4 (-1 - sqrt{41}))$$






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          Starting with $2 x^4+x^3-7 x^2-x+5$ and using the insight that $x=1$ must be a factor:



                          Divide out:



                          $${2 x^4+x^3-7 x^2-x+5 over x-1} = 2 x^3+3 x^2-4 x-5$$



                          Notice again, from the sum of coefficients that $x = -1$ is a root, so divide out again:



                          $${2 x^3+3 x^2-4 x-5 over x+1} = 2 x^2+x-5.$$



                          So:



                          $$(x-1)(x+1)(2 x^2+x-5)$$



                          Use the quadratic equation:



                          $$(x-1)(x+1)(x - 1/4 (-1 - sqrt{41}))(x+1/4 (-1 - sqrt{41}))$$






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            Starting with $2 x^4+x^3-7 x^2-x+5$ and using the insight that $x=1$ must be a factor:



                            Divide out:



                            $${2 x^4+x^3-7 x^2-x+5 over x-1} = 2 x^3+3 x^2-4 x-5$$



                            Notice again, from the sum of coefficients that $x = -1$ is a root, so divide out again:



                            $${2 x^3+3 x^2-4 x-5 over x+1} = 2 x^2+x-5.$$



                            So:



                            $$(x-1)(x+1)(2 x^2+x-5)$$



                            Use the quadratic equation:



                            $$(x-1)(x+1)(x - 1/4 (-1 - sqrt{41}))(x+1/4 (-1 - sqrt{41}))$$






                            share|cite|improve this answer









                            $endgroup$



                            Starting with $2 x^4+x^3-7 x^2-x+5$ and using the insight that $x=1$ must be a factor:



                            Divide out:



                            $${2 x^4+x^3-7 x^2-x+5 over x-1} = 2 x^3+3 x^2-4 x-5$$



                            Notice again, from the sum of coefficients that $x = -1$ is a root, so divide out again:



                            $${2 x^3+3 x^2-4 x-5 over x+1} = 2 x^2+x-5.$$



                            So:



                            $$(x-1)(x+1)(2 x^2+x-5)$$



                            Use the quadratic equation:



                            $$(x-1)(x+1)(x - 1/4 (-1 - sqrt{41}))(x+1/4 (-1 - sqrt{41}))$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 11 at 0:37









                            David G. StorkDavid G. Stork

                            10.7k31332




                            10.7k31332















                                -

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