How could you use descartes' rule and the fundamental theorem of algebra to predict the number of complex roots to a polynomial as well as find the number of possible positive and negative real roots to a polynomial? . your response must include: . a summary of descartes' rule and the fundamental theorem of algebra. this must be in your own words.. two examples of the process. provide two polynomials and predict the number of complex roots for each.. you must explain how you found the number of complex roots for each.. .

Descartes' Rule of Signs: A polynomial has no more positive roots than it has sign changes.

A corollary to the Rule of Signs: A polynomial P has no more negative roots than P(–x) has sign changes.

The Fundamental Theorem of Algebra: Any polynomial of degree n has n roots.

The Descartes' Rule of Signs gives you the possible number of positive roots (the possible number of the positive roots of a polynomial is equal to the number of sign changes in the coefficients of the terms or less than the sign changes by a multiple of 2) and corollary to the Rule of Signs gives the possible number of negative roots. Subtracting the sum of these two numbers from the degree of polynomial, you can count the number of complex roots.

Examples:

1. The polynomial has three sign changes but just one positive root (of multiplicity 3) at x = 1

2. The polynomial has two sign changes but no positive roots. On the other hand, has one sign change and g(x) has one negative root. Therefore, this polynomial has 3-1=2 complex roots (using the Fundamental Theorem of Algebra).

Descartes' Rule of Signs: A polynomial has no more positive roots than it has sign changes.

A corollary to the Rule of Signs: A polynomial P has no more negative roots than P(–x) has sign changes.

The Fundamental Theorem of Algebra: Any polynomial of degree n has n roots.

The Descartes' Rule of Signs gives you the possible number of positive roots (the possible number of the positive roots of a polynomial is equal to the number of sign changes in the coefficients of the terms or less than the sign changes by a multiple of 2) and corollary to the Rule of Signs gives the possible number of negative roots. Subtracting the sum of these two numbers from the degree of polynomial, you can count the number of complex roots.

Examples:

1. The polynomial has three sign changes but just one positive root (of multiplicity 3) at x = 1

2. The polynomial has two sign changes but no positive roots. On the other hand, has one sign change and g(x) has one negative root. Therefore, this polynomial has 3-1=2 complex roots (using the Fundamental Theorem of Algebra).

incomplete question. how do you expect anyone to you when you do not post the question in its entirety?

see explanation

step-by-step explanation:

since the sequence is geometric there is a common ratio r between consecutive terms, that is

r = 4 ÷ - 2 = - 2

r = - 8 ÷ 4 = - 2

r = 16 ÷ 8 = - 2

the next term - 13 must be incorrect as it should be - 32 for the sequence to be geometric.

the first 4 terms are therefore geometric with common ratio r = - 2