# Missing factor equal to 0 after factoring

The equation $$x^{10}+(13x-1)^{10}=0$$ has $$10$$ complex roots, $$r_i$$ for integers $$1\le i\le5$$. Find $$\sum^5_{i=1}\frac{1}{r_i\overline{r_i}}.$$

My solution was simply to factor the given polynomial into $$(r^2+(13r-1)^2)(r^8-\dots+(13r-1)^8)=0.$$ We suppose the left factor is $$0,$$ such that $$170r^2-26r+1=0.$$ By Vieta's, the product of the roots if $$1/170.$$ Since this product is the product of the conjugate pairs, we can multiply $$170\times5$$ to find our answer, and $$850$$ turns out to be correct.

However, how come we don't need to account for the right factor here? We need not consider it if it were never equal to $$0$$, so how do we prove that?

We solve equation $$x^{10}+(13x-1)^{10}=0$$.
$$\left(13-\frac{1}{x}\right)^{10}=-1$$ $$13-\frac{1}{x_k}=\cos\frac{(2k+1)\pi}{10}+i\sin\frac{(2k+1)\pi}{10},$$ $$\frac{1}{x_k}=13-\cos\frac{(2k+1)\pi}{10}-i\sin\frac{(2k+1)\pi}{10},\quad k=0\ldots9.$$ Then $$\frac{1}{x_k\bar x_k}=170-26\cos\frac{(2k+1)\pi}{10}$$ $$\sum_{k=0}^4\frac{1}{x_k\bar x_k}=850-26\left(\cos\frac{\pi}{10}+\cos\frac{3\pi}{10}+\cos\frac{7\pi}{10}+\cos\frac{9\pi}{10}\right)=850$$