How was this factoring of $1-2x-x^2$ achieved?

Question

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?

---

Answer 1

Note that: $\frac{1+x}{1-2x-x^2}=\frac{-(1+x)}{(\alpha-x)(\beta-x)}$ with $\alpha$ and $\beta$ as you have them in your quadratic formula part. Using partial fractions, we then get: $$=\frac{\frac{1}{2}}{\alpha-x}+\frac{\frac{1}{2}}{\beta-x}=\frac{\frac{1}{2}}{\frac{1}{\alpha}(1-\alpha x)}+\frac{\frac{1}{2}}{\frac{1}{\beta}(1-\beta x)}$$ Then simplifying yields what you want.

Answer 2

That's an awful lot of work for such a simple factoring problem. Take out a negative: $$-[x^2+2x-1]$$ Then complete the square inside: $$-[(x+1)^2-2]$$ Use the difference of two squares (Of course $2$ is a square...) $$-(x+1+\sqrt{2})(x+1-\sqrt{2})$$ Is this sufficient?

Answer 3

The author is using a standard technique for solving linear homogeneous recurrence relations in reverse. The recurrence $a_n=2a_{n-1}+a_{n-2}$ is the one that would generate this polynomial if you assume a solution of the form $a_n=kr^n$ where $r$ is a root of the polynomial. $\alpha$ and $\beta$ come from doing partial fractions on the $f(x)$, so factoring $1-2x-x^2=(1-\alpha x)(1-\beta x)$. The author doesn't say how they were found, but it was probably by the quadratic formula. It seems a long way around to conceal the use of the quadratic formula to me. The logic is sound, but it is a long way to go.

Answer 4

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 $x = -1-\sqrt2$).