Using Descartes theorem, find the possible numbers of positive, negative, and non-real complex numbers:
P(x) = x^3 - 5x^2 + 3x + 7
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