Descartes' rule of signs says that the number of positive roots is equal at

most to the number of sign changes in the coefficients of a polynomial, and

that the number of actual roots can only be a multiple of 2 less than this

number.

In this case there are 2 sign change, so there are 2 or 0 positive real roots

The possible number of negative roots can be investigates by considering

the sign changes in P(-x) = -x^3 - 5x^2 - 3x +7, so there is at most 1

negative root (as there is only one sign change). But there must be exactly

one such root as 1-2, 1-3, .. cannot be the number of roots.

So we conclude that there is 1 negative root, and there are either 0 or 2

positive roots, and if there are 0 positive there are 2 non-real complex roots,

and if 2 positive roots there are 0 non-real complex roots.

RonL