How to factor the polynomial $x^4-x^2 + 1$

Question

How do I factor this polynomial: $$x^4-x^2+1$$ The solution is: $$(x^2-x\sqrt{3}+1)(x^2+x\sqrt{3}+1)$$ Can you please explain what method is used there and what methods can I use generally for 4th or 5th degree polynomials?

Answer 1

Key Idea $\ $ Completing the $\rm\color{#0a0}{square}$ yields a difference of squares

$$\ \color{#c00}{2ab=c^2}\Rightarrow\! \begin{eqnarray} \overbrace{\color{#0a0}{a^2+b^2}}^{\rm incomplete\ \large \Box\!\!} &\,=\,& \overbrace{\color{#0a0}{(a+b)^2}}^{\rm\!\!\! completed\ \large \Box \!\!\!}-\ \overbrace{\color{#c00}{2ab}}^{\textstyle \color{#c00}{c^2}}\ \ \text{so factoring this} \it\text{ difference of squares}\\[.2em] &\,=\,& (a+b\color{#c00}\ \ {-\ \ \color{#c00}{c}})\ (a+b\, + \color{#c00}{c})\\ \end{eqnarray}\qquad\qquad$$

$$\begin{eqnarray} \overbrace{\color{#0a0}{x^4+1}}^{\rm incomplete\ \large \Box\!\!}\!\! -x^2 &\,=\,& \overbrace{\color{#0a0}{(x^2\!+1)^2}}^{\rm\!\!\! completed\ \large \Box \!\!\!}-\ \color{#c00}{3\, x^2}\ \ \text{so factoring this} \it\text{ difference of squares}\\[.2em] &\,=\,& (x^2+1\color{#c00}\ \ {-\ \ \color{#c00}{\sqrt3 x}})\ (x^2+1\, + \color{#c00}{\sqrt3 x})\\ \end{eqnarray}$$ Another example $$\begin{eqnarray} \overbrace{\color{#0a0}{n^4+4k^4}}^{\rm incomplete\ \large \Box}\!\! &=\,& \overbrace{\color{#0a0}{(n^2\!+2k^2)^2}}^{\rm\!\!\! completed\ \large \Box\!\!\!}\!\!-\!(\color{#c00}{2nk})^2\ \ \text{so factoring this} \it\text{ difference of squares}\\[.2em] &\,=\,& (n^2\!+2k^2\ -\,\ \color{#c00}{2nk})\,(n^2\!+2k^2+\,\color{#c00}{2nk})\\[.2em] &\,=\,&(\underbrace{(n-k)^2}_{\rm\!\!\!\!\!\!\!\!\!\!\! complete\ the\ square\!\!\!\!\!\!\!\!}\ +\ \,k^2)\ \ \underbrace{((n+k)^2}_{\rm\!\!\!\!\!\!\!\!\!\!\! complete\ the\ square\!\!\!\!\!\!\!\!\!\!\!\!\!\!} +\,k^2)\\ \end{eqnarray}$$ Another example $$\begin{eqnarray}\overbrace{3^{\large 6}+1}^{\rm incomplete\ \large \Box\!\!}\!\!-\,3^{\large 3}&=\ &\! \overbrace{(3^{\large 3}+1)^{\large 2}}^{\rm\!\!\!\!\! completed\ \large \Box\!\!\!\!\!}\!\!-\color{#c00}{3\,3^{\large 3}}\ \ \text{so, factoring this} \it\text{ difference of squares}\\[.3em] &\!\!\!=&\! (\underbrace{3^{\large 3}+1\,\ -\, \color{#c00}{3^{\large 2}}}_{\Large 19})\ (\underbrace{3^{\large 3}+1\ +\,\color{#c00}{3^2}}_{\Large 37})\\ \end{eqnarray}$$

Generally $\ a^{\large 6} + b\, a^{\large 3} + c^{\large 2}\,$ factors if $\ \color{#c00}{ b= 2c\!-\!ad^{\large 2}}$ for some $d\,$ (above is $\,a,b,c,d = 3,-1,1,1)$

$$\qquad\ \begin{eqnarray}\overbrace{a^{\large 6}+c^{\large 2}}^{\rm incomplete\ \large \Box\!\!}\!+b\,a^{\large 3}&=\ &\!\! \overbrace{(a^{\large 3}+c)^{\large 2}}^{\rm\!\!\!\!\! completed\ \large \Box\!\! \!\!\!}\!\!+\color{#c00}{\overbrace{(b-2c)}^{\!\!\large -d^{\Large 2}a}\,a^{\large 3}}\ \ \text{so, factoring this} \it\text{ difference of squares}\\[.3em] &\!\!\!=&\! ({a^{\large 3}+c\,\ -\, \color{#c00}{da^{\large 2}}})\ ({a^{\large 3}+c\ +\,\color{#c00}{da^{\large 2}}})\\ \end{eqnarray}$$

Remark $ $ This generalizes to completing a $\rm\color{#0a0}{product\ (rectangle)}$, via the AC-method, viz.

$$\begin{eqnarray} &&axy + bx + cy\, =\, d,\ \ a\ne 0\\ \overset{\times\,a}\iff\, &&\!\! (ax+c)(ay+b)\, =\, ad+bc\end{eqnarray}\qquad\qquad$$

See also here for a generalizations - (Aurifeuillian) factorizations of cyclotomic polynomials, having the form $\rm\;\Phi_n(x) = C_n(x)^2 - n\ x\ D_n(x)^2,\,$ e.g. below.

$$\begin{align} x^4 + 2^2 &= (x^2 + 2x + 2)(x^2 - 2x + 2) \\[.3em] {x^6 + 3^3} &= (x^2 + 3x + 3)(x^2 - 3x + 3)(x^3+3) \\[.3em] {x^{10} - 5^5} &= (x^4 + 5x^3 + 15x^2 + 25x + 25)(x^4 - 5x^3 + 15x^2 - 25x + 25)(x^2-5)\\[.3em] {x^{12} + 6^6} &= (x^4 + 6x^3 + 18x^2 + 36x + 36)\;(x^4 - 6x^3 + 18x^2 - 36x + 36)(x^4 + 36) \end{align}$$

Answer 2

The first to try would be to look for (rational) roots, but that is fruitless here (you need only test divisors of the constant term, but $\pm1$ is not a root).

Next you might try to factor as $(x^2+ax+b)(x^2+a'x+b')$ with integer coefficients, where once again you could conclude that $bb'=1$, so $b=b'=\pm1$. However, as the solutionm tells as, this won't work to produce integer values $a,a'$ - though if you still give it a try with $b=b'=1$, you might be led to $a,a'=\pm\sqrt 3$.

However, a "trick" works here: Try to add something simple (that is also a easy to take the square root of) in order to obtain a square. Here we have $x^4-x^2+1$, which almost looks like $x^4+2x^2+1=(x^2+1)^2$, so we have $$ x^4-x^2+1=(x^2+1)^2-3x^2 = (x^2+1)^2-(\sqrt3x)^2$$ and can apply the third binomial formula to obtain a factorization.

Answer 3

Actually you have: $$x^4-x^2+1=x^4+2x^2+1-3x^2=(x^2+1)^2-(\sqrt3 x)^2 $$

and use the identity $a^2-b^2=(a-b)(a+b)$

Answer 4

You can get it by$$x^4-x^2+1=x^4+2x^2-2x^2-x^2+1=x^4+2x^2+1-3x^2=(x^2+1)^2-(\sqrt 3x)^2.$$

Answer 5

Take $x^2=t$, and then try to factor the polynomial $t^2-t+1$.