Here's the question you clicked on:
dkoos7
What are the possible number of positive, negative, and complex zeros of f(x) = 2x3 – 5x2 – 6x + 4 ?
Descartes' Rule of Signs: do you know about it?
Yea im just confused on how to find how many numbers fall under each of the categories
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.
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.
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?
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)
Pos. Real Neg. Real Complex 2 1 0 0 1 2 the above table summarizes the two possible arrangements of the roots
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.
Thank you! your a life saver! and you know a lot about math lol
maybe the former... I don't about the latter... it's all relative you know :})