anonymous
  • anonymous
What are the possible number of positive, negative, and complex zeros of f(x) = 2x3 – 5x2 – 6x + 4 ?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Descartes' Rule of Signs: do you know about it?
anonymous
  • anonymous
Yea im just confused on how to find how many numbers fall under each of the categories
JamesJ
  • JamesJ
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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
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.
anonymous
  • anonymous
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?
anonymous
  • anonymous
Want me to continue?
anonymous
  • anonymous
Yes please if you could
anonymous
  • anonymous
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)
anonymous
  • anonymous
Pos. Real Neg. Real Complex 2 1 0 0 1 2 the above table summarizes the two possible arrangements of the roots
anonymous
  • anonymous
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.
anonymous
  • anonymous
Thank you! your a life saver! and you know a lot about math lol
anonymous
  • anonymous
maybe the former... I don't about the latter... it's all relative you know :})

Looking for something else?

Not the answer you are looking for? Search for more explanations.