Factorize in $\mathbb{R}[x]$ or $\mathbb{Q}[x]$ :


I know all roots of $P$ are complex and I don't know if there a factorization

But My try as following :



I don't know how I complete?

  • 1
    $\begingroup$ @AderinsolaJoshua Thank you sir , very difficult root here $\endgroup$ Apr 3, 2020 at 13:15

4 Answers 4


For factorisation in $\mathbb{Q}[X]$:

As you note $X^6+X^2+1$ has no real roots.

Now $X^6+X^2+1=(X^3+X+1)^2$ modulo $2$, and this cubic is irreducible modulo $2$.

Hence, if the polynomial factorises over $\mathbb{Q}$ - and so over $\mathbb{Z}$ - it factorises as the product of two cubics. But every cubic has a real root, so we can't have such a factorisation.

  • $\begingroup$ Can you explain more for the final factorization? $\endgroup$ Apr 3, 2020 at 13:16
  • $\begingroup$ I don't understand: there is no rational factorisation... $\endgroup$ Apr 3, 2020 at 15:34


Since your polynomial has real coefficient, then $$p(X)=X^6+X^2+1$$ has roots $$\{z_1,\bar z_1, z_2,\bar z_2, z_3,\bar z_3\},$$ where $\bar z$ denotes the complex conjugate of $z$. Therefore you can write $p(X)$ as $$(X^2+a_1X+b_1)(X^2+a_2X+b_2)(X^2+a_3X+b_3)$$

where $$X^2+a_iX+b_i=(X-z_i)(X-\bar z_i).$$

  • $\begingroup$ Thank you sir , but roots here very difficult $\endgroup$ Apr 3, 2020 at 13:17

Note that we can try if we have a purely imaginary root, and indeed we have this particular factorisation: $$(x^2+a)(x^4-ax^2+a^2+1)=x^6+x^2+a^3+a$$

Which fits the orignal polynomial for $\ a^3+a=1$, and this cubic has at least one real root that we can calculate by Cardan's method (I skip to the result though...)

Le set $\alpha=\sqrt[3]{108+12\sqrt{93}}\ $ then $\ a=\dfrac{\alpha}6-\dfrac 2{\alpha}$

Now we get to factorize the remaining quartic, there are formulas for this that should lead you to a factorization in $\mathbb R[x]$ in product of quadratics, but considering there are no odd powers we can try to factorize under the form:


This gives us by identification of the coefficients (note that we are interested in one possible factorization, so I just take the positive square root each time) $\begin{cases}c=\sqrt{a^2+1}\\b=\sqrt{2c+a}\end{cases}$

And there you have it:



It's root are not so difficult, because it is a reducible sextic, a bi-cubic to be précised and It's roots are

$$x_1 = -(\sqrt{2^{2/3}-(\sqrt{31}-3^{3/2})^{2/3}}i)/(2^{1/6}3^{1/4}(\sqrt{31}-3^{3/2})^{1/6})$$ $$x_2 = (\sqrt{2^{2/3}-(\sqrt{31}-3^{3/2})^{2/3}}i)/(2^{1/6}3^{1/4}(\sqrt{31}-3^{3/2})^{1/6})$$ $$x_3 = -\sqrt{\sqrt{3}(\sqrt{31}-3^{3/2})^{2/3}i+2^{2/3}\sqrt{3}i-(\sqrt{31}-3^{3/2})^{2/3}+2^{2/3}}/(2^{2/3}3^{1/4}(\sqrt{31}-3^{3/2})^{1/6})$$ $$x_4 = \sqrt{\sqrt{3}(\sqrt{31}-3^{3/2})^{2/3}i+2^{2/3}\sqrt{3}i-(\sqrt{31}-3^{3/2})^{2/3}+2^{2/3}}/(2^{2/3}3^{1/4}(\sqrt{31}-3^{3/2})^{1/6})$$ $$x_5 = -\sqrt{(-\sqrt{3}(\sqrt{31}-3^{3/2})^{2/3}i)-2^{2/3}\sqrt{3}i-(\sqrt{31}-3^{3/2})^{2/3}+2^{2/3}}/(2^{2/3}3^{1/4}(\sqrt{31}-3^{3/2})^{1/6})$$ $$x_6 = \sqrt{(-\sqrt{3}(\sqrt{31}-3^{3/2})^{2/3}i)-2^{2/3}\sqrt{3}i-(\sqrt{31}-3^{3/2})^{2/3}+2^{2/3}}/(2^{2/3}3^{1/4}(\sqrt{31}-3^{3/2})^{1/6})$$

So factoring the polynomial can be done in many ways

  1. if I factor it into two cubics, I'll need to include a square root extension from a quartic polynomial
  2. If I factor it into a quadratic and a quartic, I'll need to include a cube root extension from a sextic polynomial

Polynomial decomposition is helpful here, I factored it but the results was large and messy, so I'll post the simplest case

$x^6+x^2+1 = 0$

$x^4y-x^2y^2+1 = 0$

Where $y$ is the root of $y^3+y-1=0$


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