What are the possible number of positive, negative, and complex zeros of f(x) = 3x4 - 5x3 - x2 - 8x + 4?

I am using the Descartes Rule of Signs.

f(x) = 3x^4 - 5x³ - x² - 8x + 4

positive zero: 2 or 0.
x = + - - +
There are 2 sign changes : + to - ; - to +.

negative zero: 1.

f(-x) = 3(-x^4) - 5(-x³) - (-x²) - 8(-x) + 4
f(-x) = 3x^4 + 5x³ - x² + 8x + 4

-x = + + - + +

there is only 1 change of sign. - to +.

There are 2 or 0 positive solutions and 1 negative solution. 

The question is kind of confusing. if it's asking how many variations of answers can you give. then do 4^5 for your answer.this would be something like the first question could be a and the 2nd c or first b next c etc. if you can, ask the teacher for a better confirmation of what it's asking. if i'm assuming correctly what it's asking, it's 625 different variations.