Question

Part 1 [3 points]: What are the possible number of positive, negative, and complex zeros of f(x) = 2x3 – 5x2 + 6x – 4 ?

Answer 1

you have to count sign changes, this looks like an application of one of descartes things

Answer 2

we usually let x=+ and then -, but lets make this a tad simpler to follow, make x = +1 and we get:

f(+1) = 2 -5 +6 -4 ; and count how many times the operators change;
^ ^ ^
I count 3 changes, so we have either 3 or 1 possible positive zeros

let x=-1 to find the number of possible negative zeros (which means the odd powers change sign)

f(-1) = -2 -5 -6 -4 ; and count how many times the operators change;

I count 0 changes, so we have NO possible negative zeros

The number of possible complex zeros stems from the process of subtracting 2 from the others zeros. And since we were only able to do that once, there is at most 2 possible complex zeros