Question

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.. . .

Answer 1

Decartes' Rule of sing pretty much explains the amount of all of the positive and negative real roots or the imaginary roots for each and every function.Then, you can use Fundamental Theorem of Algebra to find the most number of total zeros Then, just subtract the number of real zeros from the total zeros to get the complex zeros.

Answer 2

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 $f(x) = x^ 3 -3 x ^2 + 3 x - 1$ has three sign changes but just one positive root (of multiplicity 3) at x = 1

2. The polynomial $g(x) = x^ 3 - x ^2 + 1$ has two sign changes but no positive roots. On the other hand, $g(-x) = -x^ 3 - x ^2 + 1$ 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).