Question

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

Answer 1

Descartes' Rule of Signs: do you know about it?

Answer 2

Yea im just confused on how to find how many numbers fall under each of the categories

Answer 3

Well, at first pass, we notice that because of the 2x^3 term, then for large negative x, f(x) is negative and for large positive x, f(x) is positive.

So the equation must have at least one zero.

I would now differentiate and see if the critical points tell you something about how the function behaves between large negative and large positive values.

Answer 4

The max number of pos. real roots is the number of sign changes of f... 2 in this case, but that number can be less by a positive even integer, i.e., 2-2=0.

Answer 5

So there are 2 or 0 pos. real roots. do you remember the next part? How to get the max number of negative real roots?

Answer 6

Want me to continue?

Answer 7

Yes please if you could

Answer 8

The max number of negative real roots is the number of sign variations of f(-x). When you replace -x for x in the rule of the function, all the terms with odd powers will change signs:

f(-x)=-2x^3-5x^2+6x+4

and now there is 1 sign variation, so the number of negative real roots is 1 (this has to be a root-- remember what james sped earlier? Odd degreed polynomial functions have at least one real root).

(put all the info together on next post)

Answer 9

Pos. Real Neg. Real Complex
2 1 0
0 1 2

the above table summarizes the two possible arrangements of the roots

Answer 10

A third degree polynomial by a corollary to the Fundamental theorem of Algebra indicates must have 3 roots, and each row shows the 3 arranged by type.

Answer 11

Thank you! your a life saver! and you know a lot about math lol

Answer 12

maybe the former... I don't about the latter... it's all relative you know :})