find all the zeros for each function.

Question

P(x)=2x^3-3x^2+3x-2

Answer 1

First look for low-hanging fruit. A little trial shows P(1)=0, so

P(x) = (x-1)(2x^2-x+2)

The quadratic formula finds the other two roots.

Since cubics are hard to solve in general, if they throw one your way, you know it must have at least one real root, and it will probably be relatively easy to find.