Question

Help please????



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

Answer 1

Positive: 3 or 1; negative: 2 or 0; complex: 4, 2, or 0
Positive: 3 or 1; negative: 3 or 1; complex: 4, 2, or 0
Positive: 4, 2, or 0; negative: 2 or 0; complex: 6, 4, 2, or 0
Positive: 2 or 0; negative: 2 or 0; complex: 6, 4, or 2

Answer 2

f(x) = -x⁶ + x⁵ - x⁴ + 4x³ - 12x² + 12

In successive terms there are 5 sign changes, so there are either 5, 3, or 1 positive real root(s).
So, from just that, it must be answer C. <==ANSWER

A.Positive: 4, 2, or 0; Negative: 2 or 0; Complex: 6, 4, 2, or 0
B.Positive: 3 or 1; Negative: 3 or 1; Complex: 4, 2, or 0
C.Positive: 5, 3, or 1; Negative: 1; Complex: 4, 2, or 0
D.Positive: 2 or 0; Negative: 2 or 0; Complex: 6, 4, or 2

If I changed the signs of terms with odd exponents, the new equation would be:

f(x) = -x⁶ - x⁵ - x⁴ - 4x³ - 12x² + 12

This equation only has 1 sign change, so there is only one negative root. This is also answer C.


If there are 5 positive roots and 1 negative, that is all 6 roots, so there are no complex roots.
If there are 3 positive roots and 1 negative, that is only 4 roots, so there are 2 complex roots.
If there is 1 positive root and 1 negative root, that is only 2 so far, so the other 4 are complex.
This is also answer C.
hope it helps :P

Answer 3

Thank you so much!! I had trouble understanding it, but I think I get it now

Answer 4

no problem im here to help

Answer 5

whoa this is a big problem :P