Proving Rational Solutions of ax^n + a0 = 0

[cooljosh2k2]

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.

 

[Mark44]

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.

 

[cooljosh2k2]

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