Proving Rational Solutions of ax^n + a0 = 0

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In summary, the problem states that if r is a rational solution of the given equation, then q must divide the coefficient of x2 (a2) and p must divide the constant term (a0). An example is provided to illustrate this theorem.
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cooljosh2k2
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Homework Statement



Suppose that r is a solution of the equation:

anxn + a(n−1)x(n−1) + . . . + a1x + a0 = 0

where the coefficients ak belongs to Z for k = 0, 1, . . . n, and n is greater or equal to 1. If r is a rational solution r = p/q, where p, q belong to Z and p and q are
coprime, show that q|an and p|a0.


The Attempt at a Solution



Im not even sure where to begin, I am so confused, what am i trying to prove? and how do i prove it, i feel like there is something missing in the question.
 
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  • #2


cooljosh2k2 said:

Homework Statement



Suppose that r is a solution of the equation:

anxn + a(n−1)x(n−1) + . . . + a1x + a0 = 0

where the coefficients ak belongs to Z for k = 0, 1, . . . n, and n is greater or equal to 1. If r is a rational solution r = p/q, where p, q belong to Z and p and q are
coprime, show that q|an and p|a0.


The Attempt at a Solution



Im not even sure where to begin, I am so confused, what am i trying to prove? and how do i prove it, i feel like there is something missing in the question.

Here's an example to help show you how this works. Here's an equation: x2 - 4x + 4 = 0.

In this equation a2, the coefficient of x2, is 1. a0 is the constant term, and is 4.

If there is a rational number r = p/q that is a solution to this equation, this theorem says that p has to divide a0, and q has to divide a2.

As it turns out, 2 is a solution, and is a rational number - i.e., 2 = 2/1. Clearly 2 divides 4, and 1 divides 1.
 
  • #3


Thanks i got it now, just got confused with the wording.
 

1. What is the general method for proving rational solutions of a polynomial equation?

The general method for proving rational solutions of a polynomial equation such as ax^n + a0 = 0 is to use the Rational Root Theorem. This theorem states that any rational solution of the equation must be a factor of the constant term a0 divided by a factor of the leading coefficient a.

2. How do you apply the Rational Root Theorem to a specific equation?

To apply the Rational Root Theorem to a specific equation, you must first identify the constant term a0 and the leading coefficient a. Then, list out all possible factors of a0 and a. These factors are the potential rational solutions of the equation. To determine which of these potential solutions are actual solutions, you can use synthetic division or plug each potential solution into the equation and see if it results in a true statement.

3. Can a polynomial equation have more than one rational solution?

Yes, a polynomial equation can have multiple rational solutions. In fact, the Fundamental Theorem of Algebra states that any polynomial equation of degree n has exactly n complex solutions, which can include both rational and irrational solutions.

4. How does the degree of the polynomial equation affect the number of rational solutions?

The degree of the polynomial equation does not necessarily determine the number of rational solutions. A polynomial equation of degree n can have anywhere from 0 to n rational solutions. However, the Rational Root Theorem can help narrow down the potential solutions and make it easier to determine the actual rational solutions.

5. Are there other methods for proving rational solutions of a polynomial equation?

Yes, there are other methods for proving rational solutions of a polynomial equation, such as the Descartes' Rule of Signs and the Factor Theorem. These methods can be used in conjunction with the Rational Root Theorem to find and prove the rational solutions of a polynomial equation.

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