Question

Using complete sentences, describe how you would analyze the zeros of the polynomial function f(x) = 7x5 + 15x4 – x3 + 4x2 – 6x – 11 using Descartes’ Rule of Signs.

Be sure to provide the answer in your explanation.

Evidently my answer is missing imaginary roots, I'm just not sure where to incorporate them.

Answer 1

Positive:
f(x)
The number of positive roots of the polynomial is either equal to the number of sign differences between consecutive non-zero coefficients, or is less than it by a multiple of 2.
There are 3 sign differences, so there are 3, 1 positive roots.
(Descartes' Rule of Signs = 3-2=1)

Negative:
f(-x)
The number of negative roots of the polynomial is either equal to the number of sign differences between consecutive non-zero coefficients of f(-x), or is less than it by a multiple of 2.
There are 2 sign differences, so there are 2, 0 negative roots.
(Descartes' Rule of Signs = 2-2=0)

Complex: Degree of polynomial is 5, so the zero for the complex number is 5.

Answer 2

@Jhannybean Luigi said you might be able to help me with this.

Answer 3

\( f(x) = 7x^5 + 15x^4 – x^3 + 4x^2 – 6x – 11 \) 3 changes of sign

\(f(-x) = -7x^5 + 15x^4 + x^3 + 4x^2 + 6x – 11 \) 2 changes of sign

Everything you wrote above is correct.

Now remember that the number of roots is the same as the degree.
Also complex roots come in pairs of complex conjugates.

Answer 4

So the complex is the imaginary? >.>

Answer 5

Yes.

Answer 6

Remember that the Descartes' Rule of Signs tells you the number of potential positive or negative roots, not a certainty. Each line in the table is a possible outcome.

Answer 7

Got it, thank you so much for your help!

Answer 8

wlcm