Solving This Polynomial

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  • #1
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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 [itex]r[/itex] is the variable I'd like to solve for and [itex]P[/itex], [itex]a[/itex] 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 [itex]r[/itex]?
 

Answers and Replies

  • #2
34,671
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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 [itex]r[/itex] is the variable I'd like to solve for and [itex]P[/itex], [itex]a[/itex] 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 [itex]r[/itex]?
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...
 
  • #3
pasmith
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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 [itex]r[/itex] is the variable I'd like to solve for and [itex]P[/itex], [itex]a[/itex] 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 [itex]r[/itex]?
There are special cases to consider.

If [itex]a = 1[/itex] and [itex]P \neq 1[/itex] there are no solutions. If [itex]a =1[/itex] and [itex]P = 1[/itex] there are infinitely many solutions.

If [itex]a \in \{2,3,4,5\}[/itex] then linear factors can be cancelled 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 [itex]a[/itex] you will have to solve a quintic, and in general it is not possible to solve quintics analytically. But the case [itex]P = 0[/itex] is trivial, as your quintic is then already factored.
 
  • #4
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Is there a numerical method to solve something like this?
 

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