# Proof of Rational Root Theorem

I want to find a nice and elegant proof of the Rational Root theorem, but I get stuck. I read some stuff on the Internet, but I have not found a complete proof of the theorem. Here's my try:

Say we have a polynomial:

$$F(x) = \sum ^{n}_{r = 0} a_{n}x^{n} = a_{n}x^{n} + a_{n - 1}x^{n-1} + ... + a_{1}x + a_{0}$$

where all of the coefficients are integers and $$a_{0}$$ does not equal zero.

I want to prove that there is a factor of $$a_{n}$$, call it q, and a factor of $$a_{0}$$, call it p, so that:

$$F(\frac{p}{q}) = 0$$

p and q are prime numbers. So I want to prove that p is a factor of $$a_{0}$$ and q is a factor of $$a_{n}$$. Next, I insert p/q into F(x):

$$F(\frac{p}{q}) = a_{n}\left(\frac{p}{q}\right)^{n} + a_{n-1}\left(\frac{p}{q}\right)^{n-1} + ... + a_{1}\left(\frac{p}{q}\right) + a_{0} = 0$$

multiplying by $$q^{n}$$ we get:

$$a_{n}p^{n} + a_{n-1}p^{n-1}q + ... + a_{1}p q^{n-1} + a_{0}q^{n} = 0$$

and factorizing out p:

$$p (a_{n}p^{n-1} - a_{n-a}p^{n-2} q - ... - a_{1} q^{n-1}) = a_{0} q^{n}$$

As p, q are prime numbers, the Greatest Common Devisor of p and q is 1, i.e. GCD(p, q) = 1. And hence, p devides $$a_{0}$$ and is then a factor of $$a_{0}$$.

As the same kind of argument applies to q being a factor of $$a_{n}$$, I thought that this proved the theorem (p/q are roots of F(x)), but no... The problem is in:

$$p (a_{n}p^{n-1} - a_{n-a}p^{n-2} q - ... - a_{1} q^{n-1}) = a_{0} q^{n}$$

how do I prove that: $$a_{n}p^{n-1} - a_{n-a}p^{n-2} q - ... - a_{1} q^{n-1}$$ is an integer?

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Paragon said:
I want to find a nice and elegant proof of the Rational Root theorem, but I get stuck. I read some stuff on the Internet, but I have not found a complete proof of the theorem. Here's my try:
Say we have a polynomial:
$$F(x) = \sum ^{n}_{r = 0} a_{n}x^{n} = a_{n}x^{n} + a_{n - 1}x^{n-1} + ... + a_{1}x + a_{0}$$
where all of the coefficients are integers and $$a_{0}$$ does not equal zero.
I want to prove that there is a factor of $$a_{n}$$, call it q, and a factor of $$a_{0}$$, call it p, so that:
$$F(\frac{p}{q}) = 0$$

You won't be able to prove that. Exercise: find a polynomial with integer coefficients which doesn't have any rational roots.

The "rational root" theorem that /I'm/ familiar with states that if p/q is a rational root of F and gcd(p, q) = 1, then q is a factor of a_n and p is a factor of a_0. (Note that this theorem doesn't guarantee that F actually has any rational roots).

how do I prove that: $$a_{n}p^{n-1} - a_{n-a}p^{n-2} q - ... - a_{1} q^{n-1}$$ is an integer?

A finite sum of integers is clearly an integer itself.

matt grime
Homework Helper
Firstly, as stated your trying to prove something that is false. There is no reason to suppose that an integer polynomial has *any* rational roots.

Then there is the issue of what a rational root might look like even if it had one: consider 2x-1, its rational root is 1/2, and not of the form p/q where p and q are primes,

However, I don't see that issue you have is at all an issue: all of the a_r are integers, p and q are integers, and multiplying and adding together integers yields integers.

Indeed, the fact that the polynomial had rational zeros was a presumption. And p, q are factors of a_0 and a_n.

The statement: "A finite sum of integers is clearly an integer itself." hit the nail =)

Thanks guys, this helped a lot!

matt grime
Homework Helper
Paragon said:
Indeed, the fact that the polynomial had rational zeros was a presumption. And p, q are factors of a_0 and a_n.

You asserted they were primes. They are at best coprime, or prime to each other.

and factorizing out p:

$$p (a_{n}p^{n-1} - a_{n-a}p^{n-2} q - ... - a_{1} q^{n-1}) = a_{0} q^{n}$$

And hence, p devides $$a_{0}$$ and is then a factor of $$a_{0}$$.

My maths isn't that strong. Im having trouble with this proof as well. Basically its the logic of saying that p divides a0 and hence is a factor. Why couldn't q be factored out of the equation (instead of p) and then divide the a0*q^n on the RHS by q and just say that q is a factor of a0 (in addition to a factor of q^n).

Lastly, I don't understand why just because p divides $$a_{0}$$ means that it is a factor of a0. The logic seems to be that just because q^n can't be factored by p therefor it implies that a0 can be factored by p.

HallsofIvy