a)If g(x)=x3-3x2+2, according to Descartess Rules of Signs, g(x) could have one positive real zero.
b)If g(x)=x3-3x2+2, according to Descartess Rules of Signs, g(x) could have one negative real zero.
Descartess Rule of Signs says that if f(x) is a polynomial with real coefficients and a constant term that is not zero, that the number of positive real zeros of f is either equal to the number
of variations in sign of f(x) or less than that number by an even integer.
g(x) has 2 variations in sign, therefore it can either have 2 positive real zeros or none (2 variations minus 2, an even integer) but it cannot have one positive real zero.
g(-x) has 1 variation in sign, therefore it has to have 1 negative real zero. (1 minus an even integer that is not 0, that is 2 gives a negative number which is not positive).
If we check the graph of g(x),
We can see it has 2 positive zeros and 1 negative zero, as Descartess Rule promised.