Question

What are the possible number of positive real, negative real, and complex zeros of
f(x) = 6x3 – 3x2 + 5x + 9?

Answer 1

easy to follow guide to descartes rule of sign is here
http://www.purplemath.com/modules/drofsign.htm

Answer 2

Use Descartes Rule of Signs
See how many times the sign changes between each term to see how many positive zeroes there might be.
3 or 1
Then plug in -x for every x and see how many times the sign changes.
1
Now, because the degree is 3, subtract these numbers from 3 to see the possibilities of complex roots.
1 or 0 complex roots.

Answer 3

You see, Descartes Rule of Signs states that the number of possible positive rational roots are calculated by the sign changes within the terms of the polynomial. Then, you plug in -x for all of the x in the polynomial so that you can calculate the number of sign changes, and therefore the number of possible negative rational roots. However, there is a rule that states that if there are 2 or more sign changes for either of these, then you have to subtract two. So, if you had 2 sign changes, you could have 2 or 0 of that type of root. Then, the degree states the number of roots and serves as a limitation. If just subtract one pair of positive and negative root possibilities from that number to get the complex roots.
Here is the sign layout of positives to negatives:
+-++
There are 3 sign changes. Also subtract 2 to get 1.
Therefore, there are either 3 or 1 possible positive rational roots. Do the same with what I said above for the negative and complex roots.