Question

What are the possible number of positive real, negative real, and complex zeros of
f(x) = 6x^3 – 3x^2 + 5x + 9?

Answer 1

Have you heard of Descartes' Rule of Signs?

Answer 2

yes but iffy on the topic

Answer 3

Well, count the sign changes of f(x). +-++ That's 2 sign changes. That means what?

Answer 4

i dont know

Answer 5

Then it appears you are not quite up to "iffy".

The degree of the polynomial dictates the total number of roots. This is degree 3, so there ARE three roots (counting any with multiplicity)

With REAL coefficients, any COMPLEX roots MUST appear in conjugate pairs. In this case, 6, 3, 5, 9 are all Real so the ONLY choices we have are:

3 Real and 0 Complex
1 Real and 2 Complex

Counting sign changes in f(x) gives us a clue about the Positive Real Roots. In this case, we have 2 sign changes. This means there are 2 positive Real roots OR none. Only 0 and 2 are choices. There cannot be 1 or 3

Counting sign changes in f(-x)
f(x) = -6x^3 – 3x^2 - 5x + 9 ---+ Just one (1)
This means there MUST be a Negative Real Root. Just one. There cannot be zero, 2, or 3.

What do you think, so far?

Answer 6

Ok this is starting to make some sense

Answer 7

Well, that's all there is to it.

It must be one of these two thing.

1 Negative Real and 2 Positive Real
or
1 Negative Real and 2 Complex

In this case, one is indeed around x = -0.8004 and the other two are Complex.