Question

What are the possible number of positive, negative, and complex zeros of f(x) = -2x3 + 5x2 - 6x + 4 ?

Answer 1

Do you know how to use Descartes' Law of Signs? Or do you need me to explain it?

Answer 2

no

Answer 3

No for which part of the question?

Answer 4

As in, do you need me to explain Descartes' Law of Signs?

Answer 5

yes i do need you to

Answer 6

OK. So DLS (Descartes' Law of Signs) basically says that the number of positive roots possible is the number of sign changes in the polynomial for f(x) minus positive integer multiples of 2, the number of negative roots possible is the number of sign changes in the polynomial for f(-x) minus positive integer multiples of 2, and the number of complex roots possible is just the difference.

Answer 7

It's easier to explain with an example, so we'll use the problem you have here. For the number of positive roots, we have 3 sign changes, since each time you hop from term to term you change the sign from + to - or vice versa. Following so far?

Answer 8

(by the way f(x) is just the polynomial)

Answer 9

Are you following my work so far?

Answer 10

I guess I'll just assume you are. So there can be 3 or 1 positive roots, since we can subtract 2 and still be positive. So for positive roots, it's 3 or 1. For negative roots, we plug in f(-x) to get\[f(-x)=2x^3+5x^2+6x+4\]Which has no sign changes so there are 0 negative roots. For complex, we know there must be 3 total roots, and so there can be either 0 or 2 complex, depending on if there are 1 or 3 positive. So our final table looks like this: