How can I solve this polynomial?

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In summary, the conversation discusses a rational function with a variable in the numerator and denominator, and the possibility of using De Moivre's Theorem to solve for the variable. However, it is mentioned that this is not a polynomial and it is not possible to solve algebraically for the variable. Special cases and possible methods for solving are also mentioned.
  • #1
iScience
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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]?
 
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  • #2
iScience said:
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
iScience said:
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 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 [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
Is there a numerical method to solve something like this?
 

1. How do you solve a polynomial equation?

To solve a polynomial equation, you can use various methods such as factoring, the quadratic formula, or long division. First, rearrange the equation so that all terms are on one side and the other side is equal to zero. Then, factor the polynomial if possible and set each factor equal to zero to find the solutions. If factoring is not possible, you can use the quadratic formula or long division to solve for the roots.

2. What is the degree of a polynomial?

The degree of a polynomial is the highest exponent in the expression. For example, the polynomial 3x^2 + 4x + 5 has a degree of 2 because the highest exponent is 2. The degree of a polynomial determines the number of possible solutions to the equation.

3. Can all polynomial equations be solved?

Yes, all polynomial equations can be solved, but the solutions may not always be real numbers. Some equations may have imaginary or complex solutions. Additionally, some equations may have multiple solutions or no solutions at all.

4. What is the difference between a root and a solution?

A root and a solution both refer to the values that make the polynomial equation true when substituted into the equation. The difference is that a root is a specific value that makes the equation equal to zero, while a solution can be any value that makes the equation true.

5. How do you check your solutions for a polynomial equation?

To check your solutions for a polynomial equation, you can substitute the values into the original equation and see if it results in a true statement. This can also be done by using a graphing calculator to graph the equation and see if the solutions intersect with the x-axis at the given values.

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