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

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 $\mathbb{C}$?

$$x^4-x^2+1= (x^2+1)^2-(\sqrt3x)^2$$ 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
See also: Show that $x^4-x^2+1$ is irreducible over $\mathbb{Q}$. — Martin Sleziak, Jun 12 '20 at 13:16

score 7 · Accepted Answer · answered Mar 25 '14 at 16:03

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.

score 2 · Answer 2 · answered Mar 25 '14 at 16:30

2

\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

egreg

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

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