Can De Moivre's Theorem Simplify Solving Complex Polynomial Equations?

[iScience]

I want to keep this question conceptual and qualitative (for now).
I have the following polynomial

$$\frac{(ar-1)(ar-2)(ar-3)(ar-4)(ar-5)}{(r-1)(r-2)(r-3)(r-4)(r-5)} = P$$
where $r$ is the variable I'd like to solve for and $P$, $a$ are just real constants.

I was wondering whether or not I could use De Moivre's Theorem here. Is there an easier way I can go about solving for $r$?

 

[Mark44]

I want to keep this question conceptual and qualitative (for now).
I have the following polynomial

$$\frac{(ar-1)(ar-2)(ar-3)(ar-4)(ar-5)}{(r-1)(r-2)(r-3)(r-4)(r-5)} = P$$
where $r$ is the variable I'd like to solve for and $P$, $a$ are just real constants.

I was wondering whether or not I could use De Moivre's Theorem here. Is there an easier way I can go about solving for $r$?

First off, that's not a polynomial, which generally looks like this: $a_nx^n + a_{n - 1}x^{n - 1} + \dots + a_2x^2 + a_1x + a_0$.
Your function is a rational function, the quotient of two polynomials. In your case, both the numerator and denominator are fifth-degree polynomials.

Regarding your question, I don't think it's possible to solve algebraically for r in the equation you posted, although you can possibly find an approximate solution using some numerical technique.

I don't see how de Moivre's Theorem is even applicable here...

 

[pasmith]

I want to keep this question conceptual and qualitative (for now).
I have the following polynomial

$$\frac{(ar-1)(ar-2)(ar-3)(ar-4)(ar-5)}{(r-1)(r-2)(r-3)(r-4)(r-5)} = P$$
where $r$ is the variable I'd like to solve for and $P$, $a$ are just real constants.

I was wondering whether or not I could use De Moivre's Theorem here. Is there an easier way I can go about solving for $r$?

There are special cases to consider.

If $a = 1$ and $P \neq 1$ there are no solutions. If $a =1$ and $P = 1$ there are infinitely many solutions.

If $a \in \{2,3,4,5\}$ then linear factors can be canceled from numerator and denominator. This reduces the problem to solving a polynomial which is of no higher degree than 4; this can always be done analytically.

For all other values of $a$ you will have to solve a quintic, and in general it is not possible to solve quintics analytically. But the case $P = 0$ is trivial, as your quintic is then already factored.

 

[iScience]

Is there a numerical method to solve something like this?