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
$$\frac{1}{x_k\bar x_k}=170-26\cos\frac{(2k+1)\pi}{10}$$
$$\sum_{k=0}^4\frac{1}{x_k\bar x_k}=850-26\left(\cos\frac{\pi}{10}+\cos\frac{3\pi}{10}+\cos\frac{7\pi}{10}+\cos\frac{9\pi}{10}\right)=850$$
