Determine how many zeros, how many real or complex, and find the roots for f (x) = x3 - 5x2 - 25x + 125.

We can write this as:-

P (x) = + x^3 - 5x^2 - 25x + 125

There are 2 changes of real sign so by Descartes Rule of signs there are either 2 positive real roots or 0 positive roots.

P (-x) = - x^3 - 5x^2 + 25x + 125

There is just one change of sign so there is exactly 1 real negative root.

125 is a multiple of 5 so By rational root theorem 5 could be a positive root.

P (5) = 125 - 125 - 125 + 125 = 0 so one zero is 5

if we divide the polynomial by (x - 5) we get the quadratic

x^2 - 25

(x + 5) (x - 5) = 0

x = 5,-5

so the roots are 5 (multiplicity 2) and - 5.

2 real positive zeroes and one real negative zero