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snapcracklepopz
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What are the possible number of positive, negative, and complex zeros of f(x) = –2x^3 – 5x^2 + 6x + 4 ? What does it mean if it has one sign change?
 one year ago
 one year ago
snapcracklepopz Group Title
What are the possible number of positive, negative, and complex zeros of f(x) = –2x^3 – 5x^2 + 6x + 4 ? What does it mean if it has one sign change?
 one year ago
 one year ago

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bloodrain Group TitleBest ResponseYou've already chosen the best response.0
i have no idea o_o
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
You need to use Descarte's rule of signs. This video might be helpful: http://www.youtube.com/watch?v=1utFwnmBOyU
 one year ago

snapcracklepopz Group TitleBest ResponseYou've already chosen the best response.0
Okay so I know that there is one possible negative root, does that mean that there would be 2 or 0 possible positive and complex zeros?
 one year ago

Reppinthe80 Group TitleBest ResponseYou've already chosen the best response.1
Recall that by Descartes' Rule of Signs, the number of sign changes in f(x) is the maximum number of positive real zeroes; the true amount of positive real zeroes is this maximum or less than that by an even number. It also states that the number of sign changes in f(x) is the number of negative real zeroes and that the true amount of negative real zeroes is this maximum or less than that by even number. With f(x) = 2x^3  5x^2  6x + 4, we see that there are 2 sign changes (one from +2x^3 to 5x^2 and a second one from 6x to +4), so Descartes' Rule of Signs guarantees that there are either 0 or 2 positive real roots. Now, we see that: f(x) = 2x^3  5x^2 + 6x + 4, which only has one sign change (from 5x^2 to +6x), so there is exactly 1 negative root (there cannot be any less than this since if we subtract the smallest even number from this, we get 1, and we cannot have a negative number of negative real zeroes). So, we have the following possibilities for real zeroes: (1) 0 positive, 1 negative (1 real zero total) (2) 2 positive, 1 negative (3 real zeroes total). Since f(x) is of degree 3, if there are not 3 real zeroes, the rest of the zeroes must be complex; thus, in case (1), there must be 2 complex zeroes (1 real zero + 2 complex zeroes = 3 zeroes total). In case (2), there are no complex zeroes. Therefore, we have the following cases for the types of zeroes: (1) 0 positive real zeroes, 1 negative real zero, and 2 complex zeroes (2) 2 positive zeroes, 1 negative zero, and 0 complex zeroes.
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
Have you looked at the video link I gave above?
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
it explains it quite well.
 one year ago

snapcracklepopz Group TitleBest ResponseYou've already chosen the best response.0
Thank you asnaseer and Reppinthe80! This really helped!
 one year ago
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