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?
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i have no idea o_o
You need to use Descarte's rule of signs. This video might be helpful: http://www.youtube.com/watch?v=1utFwnmBOyU
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?
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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.
Have you looked at the video link I gave above?
it explains it quite well.
Thank you asnaseer and Reppinthe80! This really helped!