What are the possible numbers of positive, negative, and complex zeros of f(x) = x6 − x5 − x4 + 4x3 − 12x2 + 12?

2 answers:

3 0

Using Descartes' rule of signs.
There are 4 changes of signs of coefficients:
+ to - (+x^6-x^5)
- to + (-x^4+4x^3)
+ to - (+4x^3-12x^2)
- to + (-12x^2+12)
Therefore, there are 4,2 or 0 positive roots at most. So it must be the third answer.

Akimi4 [234]3 years ago

3 0

Third option

Positive:4, 2, or 0; Negative: 2 or 0; Complex: 6, 4, 2, or 0

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