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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            Puebla de Zaragoza

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            Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc
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            Musa

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            Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo
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