# Finding roots of this polynomial

I have a polynomial, given for parameters $$x$$ in $$\mathbb{R}_+$$ and $$\alpha$$ and $$\beta$$ in $$\mathbb{R}_+^{n}$$ by :

$$P(t) = \sum\limits_{i=1}^n \left\{\left(\beta_i - \frac{\alpha_i}{x} - t\right) \prod\limits_{j = 1, j \neq i}^n (\beta_j - t)\right\}$$

How can i find roots, in function of $$\alpha,\beta,x$$ ? I have troubles factorising it further..

Edit : As a coment pointed out, since my parameter are all non-zero, this is the same a looking for roots of :

$$\sum\limits_{i=1}^n \frac{\alpha_i}{\beta_i - t} = x$$

• I am afraid that it can not be factorised further Dec 4, 2019 at 10:19
• So my only chance is numerical ?
– lrnv
Dec 4, 2019 at 10:23
• I don’t know if this helps, but as long as $\alpha_i\ne0$, the $\beta_i$ are not roots, and you can write $P(t)=\left(\prod_i(\beta_i-t)\right)\left(n-\sum_i\frac{\alpha_ix^{-1}}{\beta_i-t}\right)$. So, you are looking for roots of $\sum_i\frac{\alpha_i}{\beta_i-t}=nx$. Dec 4, 2019 at 10:44
• There is no formula for the roots for large $n$. But the second representation of the equation (as a sum of simple fractions) permits you to obtain almost any qualitative information that you may need about these roots, how they vary as functions of $x$ and easily find them numerically. Just the graph of the LHS. Dec 4, 2019 at 13:56
• If $(\beta_i)_{i=1}^n$ are pairwise distinct, then the sequence $((\beta_i-t)^{-1})_{i=1}^n$ of rational functions is $\mathbb R$-linear independent, which implies that the image of the $\mathbb R$-linear map $\mathbb R^n\mapsto\mathbb R[t]/(t^n),(\alpha_1,\dots,\alpha_n)\mapsto\sum_{i=1}^n\alpha_i\prod_{j\neq i}(\beta_i-t)$ has rank $n$, therefore surjective. In other words, if we fix a choice of pairwise distinct $(\beta_i)$ and $x\neq0$, and let $(\alpha_i)$ run through $n$-tuples of reals, then we can reach any polynomial equation. By Abel-Ruffini theorem, no radical formula exists.
– user20948
Dec 4, 2019 at 22:38