Question

What are the possible numbers of positive, negative, and complex zeros of f(x) = x6 + x5 + x4 + 4x3 − 12x2 + 12? (6 points)

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

Positive: 4, 2, or 0; negative: 2 or 0; complex: 4, 2, or 0

Positive: 2 or 0; negative: 2 or 0; complex: 4, 2, or 0

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

Answer 1

A

Explanation:

Answer 2

Answer : C

`f(x) = x^6 + x^5 + x^4 + 4x^3 - 12x^2 + 12`

For f(x), there will be 6 possible zeros

To get positive root we look at the sign changes in coefficients and count the sign changes in f(x)

We have two sign changes . Keep subtracting by 2 till we get 0

So positive roots are 2 or 0

Remaining 4 are complex. keep subtracting by 2 till we get 0

So complex roots are 4, 2 or 0

To get negative roots, we replace x with -x in f(x)

`f(-x) = (-x)^6 + (-x)^5 + (-x)^4 + 4(-x)^3 - 12(-x)^2 + 12`

`f-(x) = x^6 - x^5 + x^4 - 4x^3 - 12x^2 + 12`

To get negative root we look at the sign changes in coefficients and count the sign changes in f(-x)

We have two sign changes . Keep subtracting by 2 till we get 0

So negative roots are 2 or 0

Remaining 4 are complex. keep subtracting by 2 till we get 0

So complex roots are 4, 2 or 0

Answer is Positive: 4, 2, or 0; negative: 2 or 0; complex: 4, 2, or 0