Question

What are the possible number of positive, negative, and complex zeros of f(x) = -2x3 - 5x2 - 6x + 4 ? Please explain how to solve this equation!

Answer 1

Do you know Descartes Rule of Signs?

http://en.wikipedia.org/wiki/Descartes%27_rule_of_signs

Answer 2

Nope :C

Answer 3

I've been "taught" how to use it the exact same way but I always get it wrong!

Answer 4

If you read the link above, you'll find that the sign changes for this polynomial are
--,--,-+

for f(-x), we have
+-,-+,++

Therefore, there is on change in sign for positive and two sign changes for negative.

This indicates one positive real root and 2 or 0 negative real roots. Consequently, since the degree of the polynomial is 3, there are 3-1 or 3 - 2 complex roots.

In fact, there are 1 real root and 2 complex roots for this polynomial.

Answer 5

wait are roots the same as zeroes?

Answer 6

yes

Answer 7

So there's no negative zeroes?

Answer 8

That's right. Using Descartes, it predicts 2 or 0 negative real roots. So it was a possibility just based on the Rules of Signs. It turns out that 0 is the number for this case. DRoS is an approximation.

Answer 9

THANK YOU <3

Answer 10

NP - BTW, the real positive root is x = 0.45893.

Answer 11

Plug this into your equation, it will make the whole thing zero.