How do you factor the following polynomial?
$$1-x-x^2-x^3$$
So far, I've got numerical solutions, for example, I know $(0.543689 - x)$ is a factor , but I would really like to have an exact form for the factors, namely a form like the following:
$$(1-ax)(1-bx)(1-cx)$$
with $a,b,c \in \mathbb{C}$.
On
Deflate the polynomial by setting $y:=x+\frac13$ and you get
$$y^3+\frac23y+\frac{34}{27},$$ also, with $z=3y$,
$$z^3+6z+34=0.$$
Then let $z=u^{1/3}-2u^{-1/3}$. Plugging in the equation, and after simplification,
$$u+34-\frac8u=0$$ or $$u^2+34u-8=0.$$
The solutions of the quadratic equation are
$$\pm3\sqrt{33}-17.$$
Finally,
$$x=\frac13\left(\sqrt[3]{3\sqrt{33}-17}-\dfrac2{\sqrt[3]{3\sqrt{33}-17}}-1\right)$$ which matches the answer given by WA.
As it has only one real root (it is a decreasing function), you can find it using Cardano's method.
For this, you have to eliminate the term in $x^2$. For this, it will be simpler to change all signs. Thus \begin{align} x^3+x^2+x-1&=\Bigl(x+\frac13\Bigr)^3-\frac13x-\frac1{27}+x-1=\Bigl(x+\frac13\Bigr)^3+\frac23x-\frac{28}{27}\\[1ex] &=\Bigl(x+\frac13\Bigr)^3+\frac23\Bigl(x+\frac13\Bigr)-\frac29-\frac{28}{27}\\[1ex] &=\Bigl(x+\frac13\Bigr)^3+\frac23\Bigl(x+\frac13\Bigr)-\frac{34}{27}. \end{align} Setting $t=x+\dfrac13$, one has to solve $\;t^3+\dfrac23t-\dfrac{34}{27}=0$.
For this, set again $t=u+v$ and obtain \begin{align} t^3+\dfrac23t-\dfrac{34}{27}&=u^3+v^3+3uv(u+v)+\frac23(u+v)-\frac{34}{27}\\ &=u^3+v^3+(u+v)\Bigl(3uv+\frac23\Bigr)-\frac{34}{27}=0. \end{align} Add the condition $\;3uv=-\dfrac23$ to simplify this last equation. We obtain the system: $$\begin{cases}u^3+v^3=\dfrac{34}{27}\\[1ex]uv=-\dfrac 29\end{cases}\iff\begin{cases}u^3+v^3=\dfrac{34}{27}\\[1ex]u^3v^3=-\dfrac 8{729}\end{cases}$$ So finally, it comes down to this classic high-school problem: find two numbers, given their sum $s$ and their product $p$. The solutions consist of the roots of the quadratic equation $$z^2-sz+p=0.$$