Postby **origin415** » Sat Oct 24, 2009 2:18 am

I still don't understand this answer: using the rational roots theorem, all the roots must be a divisor of c over a divisor of 1, so both 3 and -2 should divide c, Seeing this I put -18 without much other thought, as -27 I thought should be impossible.

As further evidence, if p(-3) = p(2) = 0, then p(x) = (x + 3)(x - 2)(x - r) where r is the third root. Expanding this gives

p(x) = x^3 + (1-r)x^2 - (r + 6)x + 6r

So c again divisible by 6.

Edit: Nevermind, I looked again at the hypotheses of the rational roots theorem, and it only holds for integer coefficients, a and b could be any real. r = -9/2 produces c = -27 and satisfies all of the other stipulations.