Answer to Question #323125 in Linear Algebra for Busi

Question

Use Descartes’ Rule of Signs to determine the possible number of positive, negative and

imaginary zeros of P (x).

(i) P (x) = 2x^3+x^2-3x+6.

Answer 1

Task is to determine the possible number of positive, negative and imaginary zeros of

P(x) = 2x³ + x² - 3x + 6.

Finding Number of Positive Real Roots:

Descartes’ Rule of Signs says, that the number of positive real zeros of f(x) is either equal to the number of sign changes in f(x) or less than the number of sign changes by an even number.

Writing the given polynomial in the descending order of exponents, P(x) = 2x³ + x² - 3x + 6.

Here, the signs from left to right are:

+ + - +

Here there are two sign changes (one from + to -, and the other from - to +). So the maximum number of positive real roots is 2. Another possible number is 0.

Finding Number of Negative Real Roots:

Descartes’ Rule of Signs says, that the number of negative real zeros of f(x) is either equal to the number of sign changes in f(-x) or less than the number of sign changes by an even number.

For finding the number of negative real roots, find P(-x).

P(-x) = 2(-x)³ + (-x)² - 3(-x) + 6 = -2x³ + x² + 3x + 6.

Here, the signs from left to right are:

- + + +

There is only one sign change (from - to +) and hence the maximum number of negative real roots is 1. Since 1 cannot be decreased by an even number (because 1-2 = -1, which is negative), the actual number of negative real roots of f(x) is 1.

The number of roots of a polynomial function (considering the roots with multiplicities to be independent roots) is equal to the degree of the polynomial. In our case, number of roots is 3.

The number of imaginary roots can be calculated by substracting numbers of positive and negative real roots from the general number of roots.

Answer:

The number of negative real roots of P(x) is 1.

If the number of positive real roots is 2, then the number of imaginary roots is 3 - (1 + 2) = 0.

If the number of positive real roots is 0, then the number of imaginary roots is 3 - (1 + 0) = 2.