factor polynomial as $(1-x\phi)(1+x\phi)$ instead of $(x-\phi)(x+\phi)$

Question

I'm reading Generatingfunctionology, which is turning into a bit of an algebra review for me, and I am stumped by a step on page 9. I see the quadratic $1-x-x^2$ and I just pull the minus out ($-1(x^2+x-1)$) and plug it into the quadratic formula to get $-\frac{1\pm\sqrt5}{2}$. I normally expect polynomials to factor into groups of $(x-r)$, where $r$ is a root, so I get the factorization of $(x-\frac{1+\sqrt5}{2})(x-\frac{1-\sqrt5}{2})$, or $(x-\phi)(x+\phi)$. However, the author factors the above polynomial initially as $(1-x\phi)(1+x\phi)$, which he writes as $(1-xr_+)(1-xr_-)$. I see that he needs this form for the rest of his manipulations, but how does he obtain this form in the first place?

Answer 1

I get $\frac{-1 \pm \sqrt{5}}{2}$ as solutions. Notice that they cannot be written as $\pm \phi$ because their sum is not $0$ (neither of them are what you would usually call $\phi$ anyway, which is $\phi = \frac{1 + \sqrt{5}}{2}$).

Let $r_{\pm} = \frac{-1 \pm \sqrt{5}}{2}$. Notice that $r_{+}r_{-} = -1$, so that

$$1 - x - x^2 = (-1)(x-r_{+})(x-r_{-}) = r_{+}r_{-}(x-r_{+})(x-r_{-}) = (1+r_{+}x)(1+r_{-}x). $$

This expression is quite close to the one you provided, maybe the author defines $r_{+}$ and $r_{-}$ differently? Hope this helps.