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?