What are the possible number of positive, negative, and complex zeros of f(x) = 3x^4 – 5x^3 – x^2 – 8x + 4 ? Answer: A)Positive: 2 or 0; Negative: 2 or 0; Complex: 4 or 2 B)Positive: 4 or 2 or 0; Negative: 2 or 0; Complex: 4 or 2 or 0 C)Positive: 2 or 0; Negative: 2 or 0; Complex: 4 or 2 or 0 D)Positive: 3 or 1; Negative: 1; Complex: 2 or 0 help pleeeease:)
I think it's C
The number of positive real zeros is equal to the number of sign changes or is less than that by an even number. There are two sign changes. So the positive roots would be 2 or 0 The number of negative real zeros is equal to the number of sign changes in f(-x) so find that first.
f(-x)=3x^4+5x^3-x^2+8x+4 So 2 sign changes implies 2 or 0 negative roots
So far, it's C :D
mertsj, how can you tell if it's A or C?
It's not A because that does not allow for the possibility that there are 0 complex roots. If there are 2 positive and 2 negative (possibly) then there would be 0 complex since it has at most 4 roots.
so it's c:)
yes
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