proof of that one rational solution to a quadratic implies another

Quote:

Prove that if one solution for a quadratic equation of the form $\displaystyle x^{2}+bx+c=0$ is rational (where $\displaystyle b$ and $\displaystyle c$ are rational), then the other solution is also rational. (Use the fact that if the solutions of the equation are $\displaystyle r$ and $\displaystyle s$, then $\displaystyle x^{2}+bx+c=(x-r)(x-s)$.

For my proof, I'm hypothesizing that $\displaystyle s$ is the solution that is known to be rational.

So the rationals are closed under multiplication, addition and subtraction (and division, with nonzero divisors.) If $\displaystyle x$ is rational, then that closure ensures a rational $\displaystyle r$, but $\displaystyle x$ isn't necessarily rational. I know I'm missing something, but I've been doing math all day and I'm a little brain-burned. Can anyone nudge me past the whole non-rational $\displaystyle x$ hurdle?

Thanks! :)