Factoring $x^3-3x-1$ in terms of $\alpha$ unknown

Question

I never got a satisfactory answer here: Factoring $x^3-3x-1\in \Bbb Q[x]$ in terms of a unknown root

But all context is contained below

I want to factor $x^3-3x-1\in \Bbb Q[x]$ in terms of an unknown root $\alpha$, i.e. over $\Bbb Q(\alpha)$.

So since $\alpha$ is a root of $x^3-3x-1$ we can divide it out, obtaining:

$(x-\alpha)(x^2+\alpha x+(\alpha^2 -3))$ and obtaining $\alpha^3-3\alpha-1=0$.

Now my next thought would be to apply the quadratic formula to the quadratic factor, but this doesn't yield anything useful, atleast not seemingly, even with me manipulating what I believe is the key, that $\alpha^3-3\alpha-1=0$

We can do other manipulations:

$$x^2+\alpha x+(\alpha^2-3))=x^2+\alpha x + \frac1\alpha=0\implies \alpha x^2 +\alpha^2 x +1 = 0$$ $$\implies \frac{-\alpha^2\pm \sqrt{\alpha^4-4\alpha}}{2\alpha}=-\frac{\alpha}{2}\pm \frac{\sqrt{3\alpha^2-3\alpha}}{2\alpha}$$

Now I am lost again. I am meant to get the other two roots as:

$$\alpha^2-\alpha-2,\quad 2-\alpha^2$$

Answer 1

You can find the roots using Vieta's formulas. The roots $r_1, r_2$ satisfy $x^2+\alpha x + (\alpha^2-3)=0$, so $r_1+r_2=-\alpha$ and $r_1r_2=\alpha^2-3$. So, let's use undetermined coefficients: $$r_1=a\alpha^2+b\alpha+c$$ $$r_2=-a\alpha^2+(b-1)\alpha-c$$ where $r_2$ is chosen to satisfy the first equation. The second one gives $$r_1r_2=-a^2\alpha^4-a\alpha^3+(b^2-b-2ac)\alpha^2-c\alpha-c^2$$ $$=-a^2(3\alpha^2+\alpha)-a(3\alpha+1)+(b^2-b-2ac)\alpha^2-c\alpha-c^2$$ $$\alpha^2-3=\alpha^2(-3a^2+b^2-b-2ac)+\alpha(-a^2-3a-c)+(-a-c^2)$$ Hence we get the system $\{1=-3a^2+b^2-b-2ac, 0=-a^2-3a-c, -3=-a-c^2\}$. This has two solutions $\{a=-1,b=0,c=2\}$ and $\{a=-1, b=1, c=2\}$, the first of which corresponds to the desired roots.

Answer 2

OK, here's one way to extend your approach to reach the answer needed.

Suppose $a$ is a root. Then $a^3 = 3a+1$. This implies we can write higher powers of $a$ as quadratics of $a$: $$a^4 = 3a^2+a, \quad a^5 = a^2+9a+3, \quad a^6 = 9a^2+6a+1, \quad \dots$$

Further $x^3-3x-1 = (x-a)(x^2+ax+a^2-3) \implies$ the other roots are $r_{\pm} = \dfrac{-a\pm \sqrt{12-3a^2}}2$, as you noted.

One way from here is to write out a series expansion for the square root term and reduce all the higher order terms to quadratics to simplify. However often there is an easier way, with a bit of manipulation:

$$12-3a^2 = 16-4(1+3a)+4(3a^2+a)-15a^2+8a \\ =16+8a-15a^2-4(a^3)+4(a^4) \\= (2a^2-a-4)^2$$

So $r_{\pm} = \dfrac{-a\pm (2a^2-a-4)}2 \implies r_+ = a^2-a-2, \quad r_-= 2-a^2$.