Factoring $x^4 - x^2 + 1$
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
edited Dec 9 at 7:50
Martin Sleziak
44.7k7115270
asked Mar 25 '14 at 16:01
Morgan Wilde
271212
add a comment |
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
edited Dec 9 at 7:50
Martin Sleziak
44.7k7115270
asked Mar 25 '14 at 16:01
Morgan Wilde
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
add a comment |
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
edited Dec 9 at 7:50
Martin Sleziak
44.7k7115270
asked Mar 25 '14 at 16:01
Morgan Wilde
271212
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
edited Dec 9 at 7:50
Martin Sleziak
44.7k7115270
asked Mar 25 '14 at 16:01
Morgan Wilde
271212
edited Dec 9 at 7:50
Martin Sleziak
44.7k7115270
asked Mar 25 '14 at 16:01
Morgan Wilde
271212
edited Dec 9 at 7:50
Martin Sleziak
44.7k7115270
edited Dec 9 at 7:50
Martin Sleziak
44.7k7115270
edited Dec 9 at 7:50
Martin Sleziak
44.7k7115270
44.7k7115270
asked Mar 25 '14 at 16:01
Morgan Wilde
271212
asked Mar 25 '14 at 16:01
Morgan Wilde
271212
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
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.
add a comment |
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}
add a comment |
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2 Answers
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votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
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.
add a comment |
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.
add a comment |
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.
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.
answered Mar 25 '14 at 16:03
community wiki
MJD
add a comment |
add a comment |
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}
add a comment |
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}
add a comment |
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}
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}
answered Mar 25 '14 at 16:30
community wiki
egreg
add a comment |
add a comment |
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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