How to factor $-x^3+x^2-2$












1












$begingroup$


Is there an easy way to factor $$f(x)=-x^3+x^2-2;?$$



I have checked step-by-step calculators that all use theorems I am not very familiar with.



It doesn't seem like a sum or difference of cubes, and grouping is not an option. I tried factoring out the $-x^2$ $$f(x)=-x^2(x-1)-2$$ but that didn't get me anywhere. I am trying to find the roots to graph this function. How do I approach this?






polynomials roots graphing-functions factoring







share|cite|improve this question











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  • 3




    $begingroup$
    By inspection, $x = -1$ is a root. This is one of the four candidates suggested by the rational roots theorem.
    $endgroup$
    – T. Bongers
    Dec 11 '18 at 2:24


















1












$begingroup$


Is there an easy way to factor $$f(x)=-x^3+x^2-2;?$$



I have checked step-by-step calculators that all use theorems I am not very familiar with.



It doesn't seem like a sum or difference of cubes, and grouping is not an option. I tried factoring out the $-x^2$ $$f(x)=-x^2(x-1)-2$$ but that didn't get me anywhere. I am trying to find the roots to graph this function. How do I approach this?






polynomials roots graphing-functions factoring







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    By inspection, $x = -1$ is a root. This is one of the four candidates suggested by the rational roots theorem.
    $endgroup$
    – T. Bongers
    Dec 11 '18 at 2:24
















1












1








1





$begingroup$


Is there an easy way to factor $$f(x)=-x^3+x^2-2;?$$



I have checked step-by-step calculators that all use theorems I am not very familiar with.



It doesn't seem like a sum or difference of cubes, and grouping is not an option. I tried factoring out the $-x^2$ $$f(x)=-x^2(x-1)-2$$ but that didn't get me anywhere. I am trying to find the roots to graph this function. How do I approach this?






polynomials roots graphing-functions factoring







share|cite|improve this question











$endgroup$




Is there an easy way to factor $$f(x)=-x^3+x^2-2;?$$



I have checked step-by-step calculators that all use theorems I am not very familiar with.



It doesn't seem like a sum or difference of cubes, and grouping is not an option. I tried factoring out the $-x^2$ $$f(x)=-x^2(x-1)-2$$ but that didn't get me anywhere. I am trying to find the roots to graph this function. How do I approach this?






polynomials roots graphing-functions factoring




polynomials roots graphing-functions factoring






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '18 at 4:15









amWhy

1




1










asked Dec 11 '18 at 2:23









William AllenWilliam Allen

82




82








  • 3




    $begingroup$
    By inspection, $x = -1$ is a root. This is one of the four candidates suggested by the rational roots theorem.
    $endgroup$
    – T. Bongers
    Dec 11 '18 at 2:24
















  • 3




    $begingroup$
    By inspection, $x = -1$ is a root. This is one of the four candidates suggested by the rational roots theorem.
    $endgroup$
    – T. Bongers
    Dec 11 '18 at 2:24










3




3




$begingroup$
By inspection, $x = -1$ is a root. This is one of the four candidates suggested by the rational roots theorem.
$endgroup$
– T. Bongers
Dec 11 '18 at 2:24






$begingroup$
By inspection, $x = -1$ is a root. This is one of the four candidates suggested by the rational roots theorem.
$endgroup$
– T. Bongers
Dec 11 '18 at 2:24












2 Answers
2






active

oldest

votes


















2












$begingroup$

$$-x^3+x^2-2=-1-x^3+x^2-1$$
$$=-(1+x^3)+(x^2-1)$$
$$=-(1+x)(1-x+x^2)+(x-1)(x+1)$$
$$=(x+1)(x-1-(1-x+x^2))$$
$$=(x+1)(-x^2+2x-2)$$



Edit: At the fourth step, there is a factor of $x+1$ in both terms. So I factor this out of the expression as follows:
$$-(1+x)(1-x+x^2)+(x-1)(x+1)=(x+1)[-(1-x+x^2)]+(x+1)(x-1)$$
$$=(x+1)[-(1-x+x^2)+(x-1)]$$
$$=(x+1)(-1+x-x^2+x-1)$$
$$=(x+1)(-x^2+2x-2)$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Thank you, it will take me a bit of time to work out exactly what happened on the fourth step
    $endgroup$
    – William Allen
    Dec 11 '18 at 2:51










  • $begingroup$
    I will give an edit adding more detail.
    $endgroup$
    – gd1035
    Dec 11 '18 at 2:56



















0












$begingroup$

If you can find a root, $r$ then it will factor to $(x-r)(ax^2 + bx + c)$ and then we can you complete the square or use quadratic formula to see if $ax^2 + bx + c$ has rational roots. If it has roots $q_1$ and $q_2$ this will factor to $a(x-r)(x-q_1)(x-q_2)$



So can we find any roots. By rational root theorem, any rational roots of $-x^3+x^2 - 2=0$ will be of the form $pm 1, pm 2$.



$x=-1$ is a root as $-(-1)^3 + (-1)^2 -2 = 0$ but $x=1$ is not. $x=2$ nor is $x = -2$.



Dividing by $(x- (-1))$ or $x+1$ we get $frac {-x^3 +x^2 -2}{x + 1} = frac {-x^3}{x+1} + frac {x^2-2}{x+1}=$



$frac {-x^3- x^2}{x+1} + frac {x^2 + x^2-2}{x+1}=$



$-x^2 + frac {2x^2 - 2}{x+1} = -x^2 +frac {2x^2}{x+1} +frac {-2}{x+1}=$



$-x^2 + frac {2x^2 + 2x}{x+1} + frac {-2x - 2}{x+1}=$



$-x^2 + 2x -2$



So $-x^3 +x^2 -2 = (x+1)(-x^2 + 2x -2)$.



Can we factor $-x^3 + x^2 -2$? We could use quadratic formula or complete the square but we do know from above that $x = -1$ is the possible rational solution. And it doesn't work. ($-(-1)^2 + 2(-1)-2 = -3 ne 0$.)



So $-x^2 + 2x -2$ can not factor. (By quadratic equation the roots are $frac {-2pm sqrt{4 -8}}{-2}$ and those aren't possible. But we didn't need to do that.)



So $-x^3 +x^2 -2 = (x+1)(-x^2 + 2x -2)$. We can, for style points, bring the negative term out to get $-x^3 + x^2 -2 = -(x+1)(x^2 - 2x +2)$. But we can't factor any further.






share|cite|improve this answer









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    2 Answers
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    2 Answers
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    2












    $begingroup$

    $$-x^3+x^2-2=-1-x^3+x^2-1$$
    $$=-(1+x^3)+(x^2-1)$$
    $$=-(1+x)(1-x+x^2)+(x-1)(x+1)$$
    $$=(x+1)(x-1-(1-x+x^2))$$
    $$=(x+1)(-x^2+2x-2)$$



    Edit: At the fourth step, there is a factor of $x+1$ in both terms. So I factor this out of the expression as follows:
    $$-(1+x)(1-x+x^2)+(x-1)(x+1)=(x+1)[-(1-x+x^2)]+(x+1)(x-1)$$
    $$=(x+1)[-(1-x+x^2)+(x-1)]$$
    $$=(x+1)(-1+x-x^2+x-1)$$
    $$=(x+1)(-x^2+2x-2)$$






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Thank you, it will take me a bit of time to work out exactly what happened on the fourth step
      $endgroup$
      – William Allen
      Dec 11 '18 at 2:51










    • $begingroup$
      I will give an edit adding more detail.
      $endgroup$
      – gd1035
      Dec 11 '18 at 2:56
















    2












    $begingroup$

    $$-x^3+x^2-2=-1-x^3+x^2-1$$
    $$=-(1+x^3)+(x^2-1)$$
    $$=-(1+x)(1-x+x^2)+(x-1)(x+1)$$
    $$=(x+1)(x-1-(1-x+x^2))$$
    $$=(x+1)(-x^2+2x-2)$$



    Edit: At the fourth step, there is a factor of $x+1$ in both terms. So I factor this out of the expression as follows:
    $$-(1+x)(1-x+x^2)+(x-1)(x+1)=(x+1)[-(1-x+x^2)]+(x+1)(x-1)$$
    $$=(x+1)[-(1-x+x^2)+(x-1)]$$
    $$=(x+1)(-1+x-x^2+x-1)$$
    $$=(x+1)(-x^2+2x-2)$$






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Thank you, it will take me a bit of time to work out exactly what happened on the fourth step
      $endgroup$
      – William Allen
      Dec 11 '18 at 2:51










    • $begingroup$
      I will give an edit adding more detail.
      $endgroup$
      – gd1035
      Dec 11 '18 at 2:56














    2












    2








    2





    $begingroup$

    $$-x^3+x^2-2=-1-x^3+x^2-1$$
    $$=-(1+x^3)+(x^2-1)$$
    $$=-(1+x)(1-x+x^2)+(x-1)(x+1)$$
    $$=(x+1)(x-1-(1-x+x^2))$$
    $$=(x+1)(-x^2+2x-2)$$



    Edit: At the fourth step, there is a factor of $x+1$ in both terms. So I factor this out of the expression as follows:
    $$-(1+x)(1-x+x^2)+(x-1)(x+1)=(x+1)[-(1-x+x^2)]+(x+1)(x-1)$$
    $$=(x+1)[-(1-x+x^2)+(x-1)]$$
    $$=(x+1)(-1+x-x^2+x-1)$$
    $$=(x+1)(-x^2+2x-2)$$






    share|cite|improve this answer











    $endgroup$



    $$-x^3+x^2-2=-1-x^3+x^2-1$$
    $$=-(1+x^3)+(x^2-1)$$
    $$=-(1+x)(1-x+x^2)+(x-1)(x+1)$$
    $$=(x+1)(x-1-(1-x+x^2))$$
    $$=(x+1)(-x^2+2x-2)$$



    Edit: At the fourth step, there is a factor of $x+1$ in both terms. So I factor this out of the expression as follows:
    $$-(1+x)(1-x+x^2)+(x-1)(x+1)=(x+1)[-(1-x+x^2)]+(x+1)(x-1)$$
    $$=(x+1)[-(1-x+x^2)+(x-1)]$$
    $$=(x+1)(-1+x-x^2+x-1)$$
    $$=(x+1)(-x^2+2x-2)$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 11 '18 at 2:59

























    answered Dec 11 '18 at 2:35









    gd1035gd1035

    4721210




    4721210








    • 1




      $begingroup$
      Thank you, it will take me a bit of time to work out exactly what happened on the fourth step
      $endgroup$
      – William Allen
      Dec 11 '18 at 2:51










    • $begingroup$
      I will give an edit adding more detail.
      $endgroup$
      – gd1035
      Dec 11 '18 at 2:56














    • 1




      $begingroup$
      Thank you, it will take me a bit of time to work out exactly what happened on the fourth step
      $endgroup$
      – William Allen
      Dec 11 '18 at 2:51










    • $begingroup$
      I will give an edit adding more detail.
      $endgroup$
      – gd1035
      Dec 11 '18 at 2:56








    1




    1




    $begingroup$
    Thank you, it will take me a bit of time to work out exactly what happened on the fourth step
    $endgroup$
    – William Allen
    Dec 11 '18 at 2:51




    $begingroup$
    Thank you, it will take me a bit of time to work out exactly what happened on the fourth step
    $endgroup$
    – William Allen
    Dec 11 '18 at 2:51












    $begingroup$
    I will give an edit adding more detail.
    $endgroup$
    – gd1035
    Dec 11 '18 at 2:56




    $begingroup$
    I will give an edit adding more detail.
    $endgroup$
    – gd1035
    Dec 11 '18 at 2:56











    0












    $begingroup$

    If you can find a root, $r$ then it will factor to $(x-r)(ax^2 + bx + c)$ and then we can you complete the square or use quadratic formula to see if $ax^2 + bx + c$ has rational roots. If it has roots $q_1$ and $q_2$ this will factor to $a(x-r)(x-q_1)(x-q_2)$



    So can we find any roots. By rational root theorem, any rational roots of $-x^3+x^2 - 2=0$ will be of the form $pm 1, pm 2$.



    $x=-1$ is a root as $-(-1)^3 + (-1)^2 -2 = 0$ but $x=1$ is not. $x=2$ nor is $x = -2$.



    Dividing by $(x- (-1))$ or $x+1$ we get $frac {-x^3 +x^2 -2}{x + 1} = frac {-x^3}{x+1} + frac {x^2-2}{x+1}=$



    $frac {-x^3- x^2}{x+1} + frac {x^2 + x^2-2}{x+1}=$



    $-x^2 + frac {2x^2 - 2}{x+1} = -x^2 +frac {2x^2}{x+1} +frac {-2}{x+1}=$



    $-x^2 + frac {2x^2 + 2x}{x+1} + frac {-2x - 2}{x+1}=$



    $-x^2 + 2x -2$



    So $-x^3 +x^2 -2 = (x+1)(-x^2 + 2x -2)$.



    Can we factor $-x^3 + x^2 -2$? We could use quadratic formula or complete the square but we do know from above that $x = -1$ is the possible rational solution. And it doesn't work. ($-(-1)^2 + 2(-1)-2 = -3 ne 0$.)



    So $-x^2 + 2x -2$ can not factor. (By quadratic equation the roots are $frac {-2pm sqrt{4 -8}}{-2}$ and those aren't possible. But we didn't need to do that.)



    So $-x^3 +x^2 -2 = (x+1)(-x^2 + 2x -2)$. We can, for style points, bring the negative term out to get $-x^3 + x^2 -2 = -(x+1)(x^2 - 2x +2)$. But we can't factor any further.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      If you can find a root, $r$ then it will factor to $(x-r)(ax^2 + bx + c)$ and then we can you complete the square or use quadratic formula to see if $ax^2 + bx + c$ has rational roots. If it has roots $q_1$ and $q_2$ this will factor to $a(x-r)(x-q_1)(x-q_2)$



      So can we find any roots. By rational root theorem, any rational roots of $-x^3+x^2 - 2=0$ will be of the form $pm 1, pm 2$.



      $x=-1$ is a root as $-(-1)^3 + (-1)^2 -2 = 0$ but $x=1$ is not. $x=2$ nor is $x = -2$.



      Dividing by $(x- (-1))$ or $x+1$ we get $frac {-x^3 +x^2 -2}{x + 1} = frac {-x^3}{x+1} + frac {x^2-2}{x+1}=$



      $frac {-x^3- x^2}{x+1} + frac {x^2 + x^2-2}{x+1}=$



      $-x^2 + frac {2x^2 - 2}{x+1} = -x^2 +frac {2x^2}{x+1} +frac {-2}{x+1}=$



      $-x^2 + frac {2x^2 + 2x}{x+1} + frac {-2x - 2}{x+1}=$



      $-x^2 + 2x -2$



      So $-x^3 +x^2 -2 = (x+1)(-x^2 + 2x -2)$.



      Can we factor $-x^3 + x^2 -2$? We could use quadratic formula or complete the square but we do know from above that $x = -1$ is the possible rational solution. And it doesn't work. ($-(-1)^2 + 2(-1)-2 = -3 ne 0$.)



      So $-x^2 + 2x -2$ can not factor. (By quadratic equation the roots are $frac {-2pm sqrt{4 -8}}{-2}$ and those aren't possible. But we didn't need to do that.)



      So $-x^3 +x^2 -2 = (x+1)(-x^2 + 2x -2)$. We can, for style points, bring the negative term out to get $-x^3 + x^2 -2 = -(x+1)(x^2 - 2x +2)$. But we can't factor any further.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        If you can find a root, $r$ then it will factor to $(x-r)(ax^2 + bx + c)$ and then we can you complete the square or use quadratic formula to see if $ax^2 + bx + c$ has rational roots. If it has roots $q_1$ and $q_2$ this will factor to $a(x-r)(x-q_1)(x-q_2)$



        So can we find any roots. By rational root theorem, any rational roots of $-x^3+x^2 - 2=0$ will be of the form $pm 1, pm 2$.



        $x=-1$ is a root as $-(-1)^3 + (-1)^2 -2 = 0$ but $x=1$ is not. $x=2$ nor is $x = -2$.



        Dividing by $(x- (-1))$ or $x+1$ we get $frac {-x^3 +x^2 -2}{x + 1} = frac {-x^3}{x+1} + frac {x^2-2}{x+1}=$



        $frac {-x^3- x^2}{x+1} + frac {x^2 + x^2-2}{x+1}=$



        $-x^2 + frac {2x^2 - 2}{x+1} = -x^2 +frac {2x^2}{x+1} +frac {-2}{x+1}=$



        $-x^2 + frac {2x^2 + 2x}{x+1} + frac {-2x - 2}{x+1}=$



        $-x^2 + 2x -2$



        So $-x^3 +x^2 -2 = (x+1)(-x^2 + 2x -2)$.



        Can we factor $-x^3 + x^2 -2$? We could use quadratic formula or complete the square but we do know from above that $x = -1$ is the possible rational solution. And it doesn't work. ($-(-1)^2 + 2(-1)-2 = -3 ne 0$.)



        So $-x^2 + 2x -2$ can not factor. (By quadratic equation the roots are $frac {-2pm sqrt{4 -8}}{-2}$ and those aren't possible. But we didn't need to do that.)



        So $-x^3 +x^2 -2 = (x+1)(-x^2 + 2x -2)$. We can, for style points, bring the negative term out to get $-x^3 + x^2 -2 = -(x+1)(x^2 - 2x +2)$. But we can't factor any further.






        share|cite|improve this answer









        $endgroup$



        If you can find a root, $r$ then it will factor to $(x-r)(ax^2 + bx + c)$ and then we can you complete the square or use quadratic formula to see if $ax^2 + bx + c$ has rational roots. If it has roots $q_1$ and $q_2$ this will factor to $a(x-r)(x-q_1)(x-q_2)$



        So can we find any roots. By rational root theorem, any rational roots of $-x^3+x^2 - 2=0$ will be of the form $pm 1, pm 2$.



        $x=-1$ is a root as $-(-1)^3 + (-1)^2 -2 = 0$ but $x=1$ is not. $x=2$ nor is $x = -2$.



        Dividing by $(x- (-1))$ or $x+1$ we get $frac {-x^3 +x^2 -2}{x + 1} = frac {-x^3}{x+1} + frac {x^2-2}{x+1}=$



        $frac {-x^3- x^2}{x+1} + frac {x^2 + x^2-2}{x+1}=$



        $-x^2 + frac {2x^2 - 2}{x+1} = -x^2 +frac {2x^2}{x+1} +frac {-2}{x+1}=$



        $-x^2 + frac {2x^2 + 2x}{x+1} + frac {-2x - 2}{x+1}=$



        $-x^2 + 2x -2$



        So $-x^3 +x^2 -2 = (x+1)(-x^2 + 2x -2)$.



        Can we factor $-x^3 + x^2 -2$? We could use quadratic formula or complete the square but we do know from above that $x = -1$ is the possible rational solution. And it doesn't work. ($-(-1)^2 + 2(-1)-2 = -3 ne 0$.)



        So $-x^2 + 2x -2$ can not factor. (By quadratic equation the roots are $frac {-2pm sqrt{4 -8}}{-2}$ and those aren't possible. But we didn't need to do that.)



        So $-x^3 +x^2 -2 = (x+1)(-x^2 + 2x -2)$. We can, for style points, bring the negative term out to get $-x^3 + x^2 -2 = -(x+1)(x^2 - 2x +2)$. But we can't factor any further.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 11 '18 at 4:38









        fleabloodfleablood

        71.2k22686




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            Plaza Victoria

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            Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
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            In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

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            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
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            How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

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            1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
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