Finding the root of an indefinite polynomial

Question

$0 = (a-n) x^{n-1} + ax^{n-2} + ax^{n-3} + \cdots + ax + a$ What is $x$ in terms of $a$ and $n$?

I don't even know what this form of polynomial is called.

EDIT: In the required algorithm, $a$ and $n$ are never equal to $0$, and $n≥a+2$

EDIT2: the value of X is then used in the following formula

$ F(z)=\frac{nx^{z-1}}{\sum_{i=1}^n x^{i-1}} $

Answer 1

If we do a bit of reorganizing we can get $$nx^{n-1}=a\left(\sum_{k=0}^{n-1}x^k\right)$$ Using the partial sum formula for geometric series we get $$nx^{n-1} =a\frac{1-x^{n}}{1-x}$$ Doing some work with it... $$(1-x)nx^{n-1}=a(1-x^n)$$ $$a(1-x^{n})-(1-x)nx^{n-1}=0$$ $$(n-a)x^n-nx^{n-1}+a=0$$ At this point there doesn't seem to be any known general factorizations, but this form definitely looks less intimidating. I find it a little odd however that we managed to go from a $n-1$ degree polynomial to a $n$ degree polynomial, so please point out any mistakes (I feel like you would just get some extraneuos solutions). Note that if we assume that $a\neq 1$ we find that $x=1$ is said extraneous solution.

Answer 2

This is a polynomial of degree $n-1$ in $x$ (unless $a=n$). If, for example, $a$ is an integer and $p$ is a prime such that $p$ but not $p^2$ divides $a$ while $p$ does not divide $n$, then Eisenstein's criterion says it is irreducible over the rationals. In general, for $n \ge 6$ we would expect it not to have roots expressible in radicals.

EDIT: For example, if $n=6$ and $a=2$ the polynomial $-4 x^5 + 2 x^4 + 2 x^3 + 2 x^2 + 2 x + 2$ has Galois group $S_5$, so its roots are not expressible in radicals.

EDIT: Multiplying by $(1-x)/a$, we get the polynomial $(n/a-1) x^n - (n/a) x^{n-1} + 1$. With $$s = \left(\frac{n}{a}-1\right)^{-1/n} x^{-1},\ c = \left(\frac{n}{a}-1\right)^{1/n} \frac{n}{n-a}$$ the equation becomes $s^n - c + 1 = 0$. This has series solutions in powers of $c$:

$$ s = \omega - \sum_{k=1}^\infty \frac{\prod_{j=1}^{k-1} (jn-k-1)}{k! n^k} \;\omega^{k+1} c^k $$ where $\omega$ is an $n$'th root of $-1$. I believe this will converge for small enough $c$ (corresponding to large $a$).

Answer 3

If $a=0$ then $x=0$.

If $a\ne0$ then we can divide both sides by $a$ to get

\begin{equation} \left( 1-\dfrac{n}{a}\right)x^{n-1}+x^{n-2}+x^{n-3}+\cdots+x+1=0 \end{equation}

If $a=1$ then the equation becomes

\begin{equation} \left( 1-n\right)x^{n-1}+x^{n-2}+x^{n-3}+\cdots+x+1=0 \end{equation}

for which $x=1$ is a solution.

If $a\ne1$ then the $n$ in the coefficient is superfluous and the equation may as well be

\begin{equation} cx^{n-1}+x^{n-2}+x^{n-3}+\cdots+x+1=0 \text{ for }a\ne1 \end{equation}

which becomes

\begin{equation} cx^{n-1}+\dfrac{x^{n-1}-1}{x-1}=0 \text{ for }x\ne1 \end{equation}

which reduces to

\begin{equation} cx^n+(1-c)x^{n-1}-1=0 \text{ provided that }x\ne1 \end{equation}

Note that $x=1$ is a solution to the original equation when $a=1$ so the question becomes "What are the solutions to the $c$ equation other than $x=1$?

I would suggest investigating what Vieta's formulas tell us.

Vieta's Formulas