Factoring $x^4 - x^2 + 1$












2














I'm interesting in finding the possible quadratic factorization of this polynomial: $x^4 - x^2 + 1$. My first idea was to do long division by $x^2+1$, but I did get a remainder, so I presume this doesn't have a quadratic factorization over $mathbb{R}$.



How should I go about factoring the polynomial over $mathbb{C}$?






polynomials complex-numbers factoring







share|cite|improve this question




















  • 2




    Try to do $X:=x^2$
    – T_O
    Mar 25 '14 at 16:03






  • 2




    $$x^4-x^2+1= (x^2+1)^2-(sqrt3x)^2$$ or $$(x^2-1)^2-(ix)^2$$
    – lab bhattacharjee
    Mar 25 '14 at 16:03












  • Yes, this is the right way. $(x^2-sqrt{3}x+1)(x^2+sqrt{3}x+1)$
    – kmitov
    Mar 25 '14 at 16:07






  • 1




    Your equation is a particular biquadratic equation.
    – Américo Tavares
    Mar 25 '14 at 16:18


















2














I'm interesting in finding the possible quadratic factorization of this polynomial: $x^4 - x^2 + 1$. My first idea was to do long division by $x^2+1$, but I did get a remainder, so I presume this doesn't have a quadratic factorization over $mathbb{R}$.



How should I go about factoring the polynomial over $mathbb{C}$?






polynomials complex-numbers factoring







share|cite|improve this question




















  • 2




    Try to do $X:=x^2$
    – T_O
    Mar 25 '14 at 16:03






  • 2




    $$x^4-x^2+1= (x^2+1)^2-(sqrt3x)^2$$ or $$(x^2-1)^2-(ix)^2$$
    – lab bhattacharjee
    Mar 25 '14 at 16:03












  • Yes, this is the right way. $(x^2-sqrt{3}x+1)(x^2+sqrt{3}x+1)$
    – kmitov
    Mar 25 '14 at 16:07






  • 1




    Your equation is a particular biquadratic equation.
    – Américo Tavares
    Mar 25 '14 at 16:18
















2












2








2







I'm interesting in finding the possible quadratic factorization of this polynomial: $x^4 - x^2 + 1$. My first idea was to do long division by $x^2+1$, but I did get a remainder, so I presume this doesn't have a quadratic factorization over $mathbb{R}$.



How should I go about factoring the polynomial over $mathbb{C}$?






polynomials complex-numbers factoring







share|cite|improve this question















I'm interesting in finding the possible quadratic factorization of this polynomial: $x^4 - x^2 + 1$. My first idea was to do long division by $x^2+1$, but I did get a remainder, so I presume this doesn't have a quadratic factorization over $mathbb{R}$.



How should I go about factoring the polynomial over $mathbb{C}$?






polynomials complex-numbers factoring




polynomials complex-numbers factoring






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 at 7:50









Martin Sleziak

44.7k7115270




44.7k7115270










asked Mar 25 '14 at 16:01









Morgan Wilde

271212




271212








  • 2




    Try to do $X:=x^2$
    – T_O
    Mar 25 '14 at 16:03






  • 2




    $$x^4-x^2+1= (x^2+1)^2-(sqrt3x)^2$$ or $$(x^2-1)^2-(ix)^2$$
    – lab bhattacharjee
    Mar 25 '14 at 16:03












  • Yes, this is the right way. $(x^2-sqrt{3}x+1)(x^2+sqrt{3}x+1)$
    – kmitov
    Mar 25 '14 at 16:07






  • 1




    Your equation is a particular biquadratic equation.
    – Américo Tavares
    Mar 25 '14 at 16:18
















  • 2




    Try to do $X:=x^2$
    – T_O
    Mar 25 '14 at 16:03






  • 2




    $$x^4-x^2+1= (x^2+1)^2-(sqrt3x)^2$$ or $$(x^2-1)^2-(ix)^2$$
    – lab bhattacharjee
    Mar 25 '14 at 16:03












  • Yes, this is the right way. $(x^2-sqrt{3}x+1)(x^2+sqrt{3}x+1)$
    – kmitov
    Mar 25 '14 at 16:07






  • 1




    Your equation is a particular biquadratic equation.
    – Américo Tavares
    Mar 25 '14 at 16:18










2




2




Try to do $X:=x^2$
– T_O
Mar 25 '14 at 16:03




Try to do $X:=x^2$
– T_O
Mar 25 '14 at 16:03




2




2




$$x^4-x^2+1= (x^2+1)^2-(sqrt3x)^2$$ or $$(x^2-1)^2-(ix)^2$$
– lab bhattacharjee
Mar 25 '14 at 16:03






$$x^4-x^2+1= (x^2+1)^2-(sqrt3x)^2$$ or $$(x^2-1)^2-(ix)^2$$
– lab bhattacharjee
Mar 25 '14 at 16:03














Yes, this is the right way. $(x^2-sqrt{3}x+1)(x^2+sqrt{3}x+1)$
– kmitov
Mar 25 '14 at 16:07




Yes, this is the right way. $(x^2-sqrt{3}x+1)(x^2+sqrt{3}x+1)$
– kmitov
Mar 25 '14 at 16:07




1




1




Your equation is a particular biquadratic equation.
– Américo Tavares
Mar 25 '14 at 16:18






Your equation is a particular biquadratic equation.
– Américo Tavares
Mar 25 '14 at 16:18












2 Answers
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Substitute $y = x^2$; then you get $$y^2 - y + 1$$ which you can factor in the usual way (for example, with the quadratic formula) into something of the form $$(y-a)(y-b).$$ Then put back $x^2$: $$(x^2-a)(x^2-b)$$ and each of the two terms is a difference of squares.






share|cite|improve this answer































    1














    begin{align}
    x^4-x^2+1&=(x^2+1)^2-3x^2\
    &=(x^2-sqrt{3},x+1)(x^2+sqrt{3},x+1)
    end{align}






    share|cite|improve this answer























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      2 Answers
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      2 Answers
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      6














      Substitute $y = x^2$; then you get $$y^2 - y + 1$$ which you can factor in the usual way (for example, with the quadratic formula) into something of the form $$(y-a)(y-b).$$ Then put back $x^2$: $$(x^2-a)(x^2-b)$$ and each of the two terms is a difference of squares.






      share|cite|improve this answer




























        6














        Substitute $y = x^2$; then you get $$y^2 - y + 1$$ which you can factor in the usual way (for example, with the quadratic formula) into something of the form $$(y-a)(y-b).$$ Then put back $x^2$: $$(x^2-a)(x^2-b)$$ and each of the two terms is a difference of squares.






        share|cite|improve this answer


























          6












          6








          6






          Substitute $y = x^2$; then you get $$y^2 - y + 1$$ which you can factor in the usual way (for example, with the quadratic formula) into something of the form $$(y-a)(y-b).$$ Then put back $x^2$: $$(x^2-a)(x^2-b)$$ and each of the two terms is a difference of squares.






          share|cite|improve this answer














          Substitute $y = x^2$; then you get $$y^2 - y + 1$$ which you can factor in the usual way (for example, with the quadratic formula) into something of the form $$(y-a)(y-b).$$ Then put back $x^2$: $$(x^2-a)(x^2-b)$$ and each of the two terms is a difference of squares.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          answered Mar 25 '14 at 16:03


























          community wiki





          MJD
























              1














              begin{align}
              x^4-x^2+1&=(x^2+1)^2-3x^2\
              &=(x^2-sqrt{3},x+1)(x^2+sqrt{3},x+1)
              end{align}






              share|cite|improve this answer




























                1














                begin{align}
                x^4-x^2+1&=(x^2+1)^2-3x^2\
                &=(x^2-sqrt{3},x+1)(x^2+sqrt{3},x+1)
                end{align}






                share|cite|improve this answer


























                  1












                  1








                  1






                  begin{align}
                  x^4-x^2+1&=(x^2+1)^2-3x^2\
                  &=(x^2-sqrt{3},x+1)(x^2+sqrt{3},x+1)
                  end{align}






                  share|cite|improve this answer














                  begin{align}
                  x^4-x^2+1&=(x^2+1)^2-3x^2\
                  &=(x^2-sqrt{3},x+1)(x^2+sqrt{3},x+1)
                  end{align}







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  answered Mar 25 '14 at 16:30


























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