<s>[INST] How was this factoring of $1-2x-x^2$ achieved? If I factor $1-2x-x^2$ using the quadratic formula I get
$$x=\frac{2\pm \sqrt{4-4(-1)(1)}}{2(-1)}$$
$$x=\frac{2\pm \sqrt{8}}{-2}$$
$$x=-1 \pm \sqrt{2}$$
Let $\alpha = -1 +\sqrt{2}$ and $\beta=-1-\sqrt{2}$. 
So $1-2x-x^2=-(x-\alpha)(x-\beta)$.
In the image below, where did $1-\alpha x$ and $1-\beta x$ come from?


 [/INST] Your $\alpha$ and $\beta$ are not the same as the $\alpha$ and $\beta$
in the image you copied.
In the image, they are factoring
$$1−2x−x^2 = -\left(\frac 1\alpha - x\right)\left(\frac 1\beta - x\right).$$
You can confirm that using their values
$\alpha = 1 + \sqrt2$ and $\beta = 1-\sqrt2$,
we have
\begin{align}
\frac 1\alpha &= \frac 1{1 + \sqrt2} = -1+\sqrt2, \\
\frac 1\beta  &= \frac 1{1 - \sqrt2} = -1-\sqrt2, \\
\end{align}
which matches the factorization you found
(roots at $x = -1+\sqrt2$ and $x = -1-\sqrt2$).
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