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dkoos7
Group Title
What are the possible number of positive, negative, and complex zeros of f(x) = 2x3 – 5x2 – 6x + 4 ?
 3 years ago
 3 years ago
dkoos7 Group Title
What are the possible number of positive, negative, and complex zeros of f(x) = 2x3 – 5x2 – 6x + 4 ?
 3 years ago
 3 years ago

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Mandolino Group TitleBest ResponseYou've already chosen the best response.2
Descartes' Rule of Signs: do you know about it?
 3 years ago

dkoos7 Group TitleBest ResponseYou've already chosen the best response.0
Yea im just confused on how to find how many numbers fall under each of the categories
 3 years ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.0
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.
 3 years ago

Mandolino Group TitleBest ResponseYou've already chosen the best response.2
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., 22=0.
 3 years ago

Mandolino Group TitleBest ResponseYou've already chosen the best response.2
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?
 3 years ago

Mandolino Group TitleBest ResponseYou've already chosen the best response.2
Want me to continue?
 3 years ago

dkoos7 Group TitleBest ResponseYou've already chosen the best response.0
Yes please if you could
 3 years ago

Mandolino Group TitleBest ResponseYou've already chosen the best response.2
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^35x^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)
 3 years ago

Mandolino Group TitleBest ResponseYou've already chosen the best response.2
Pos. Real Neg. Real Complex 2 1 0 0 1 2 the above table summarizes the two possible arrangements of the roots
 3 years ago

Mandolino Group TitleBest ResponseYou've already chosen the best response.2
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.
 3 years ago

dkoos7 Group TitleBest ResponseYou've already chosen the best response.0
Thank you! your a life saver! and you know a lot about math lol
 3 years ago

Mandolino Group TitleBest ResponseYou've already chosen the best response.2
maybe the former... I don't about the latter... it's all relative you know :})
 3 years ago
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