It isn’t the right answer to the question as posed.  However it looks like (but isn’t necesarrily) the right answer to a very similar question, a third variant.

I would be very interested in seeing your working if you have time to write it up.

This is what I think you missed in the problem statement:

[spoiler]Although the co-efficients are constrained to be fractions there are no constraints on the test numbers.[/spoiler]

The actual answer relies on a neat bit of theory I intend to explain in a follow up post.  It relates to: [spoiler]algebraic and trancendental numbers[/spoiler]

This is my analysis of the third variant, I will write it up properly in a follow up post.

[spoiler]If we constrain the test numbers to be integers or fractions then there is no solution.  If we knew the degree of the polynomial, n, then we could do it in n+1 test numbers. Otherwise it takes an infinite number of guesses; there is a proof by contradiction as follows:  Suppose the polynomial is p(x) and we can identify it with the test numbers t0, t1, t2, … , tm.  Now put q(x) = (x-t0)(x-t1)…(x-tm).  Clearly q(ti) will be zero for each of the test numbers.  Put r(x) = q(x) + p(x).  r(ti) = p(ti) for each of the test numbers so we cannot distinguish between r(x) and p(x).  Therefore we have a proof by contradiction.[/spoiler]

Thanks czarandy, that was fun!