# 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 $\mathbb{C}$?

• Try to do $X:=x^2$ – T_O 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
• Your equation is a particular biquadratic equation. – Américo Tavares Mar 25 '14 at 16:18
• – Martin Sleziak Jun 12 '20 at 13:16

## 2 Answers

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.

\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}