Factorize : $P(x)=x^6+x^2+1$

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

$$P(x)=x^6+x^2+1$$

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

But My try as following :

$$P(x)=x^6+x^2+1=(x^3)^2+1+2x^3-2x^3+x^2$$

$$=(x^3+1)^2+x^2-2x^3$$

I don't know how I complete?

• @AderinsolaJoshua Thank you sir , very difficult root here Apr 3, 2020 at 13:15

4 Answers

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.

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

Hint

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

• Thank you sir , but roots here very difficult 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:

$$(x^2+bx+c)(x^2-bx+c)=x^4+(2c-b^2)x^2+c^2$$

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:

$$x^6+x^2+1=(x^2+a)(x^2+bx+c)(x^2-bx+c)$$

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 $$y^3+y-1=0$$