Descartes' Rule of Signs?

Question

I do not understand Descartes' Rule of Signs, so please help. I am not necessarily looking for the specific answer, but some direction on how to do these types of problems.

Use Descartes' Rule of Signs to determine how many positive and how many negative real zeros the polynomial can have. Then determine the possible total number of real zeros:

P(x) = 2x^3 - x^2 + 4x -7
and
P(x) = x^8 - x^5 + x^4 - x^3 + x^2 - x + 1

Answer 1

Dascartes rule of signs says that the number of positive roots of a polynomial (with real coefficients and ordered in descending powers) is equal to the number of sign changes between consecutive coefficients or is less than that by an exact multiple of 2 (since double roots are counted twice) Negative roots are counted by changingthe signs of the coefficients of all odd powers of the variable and counting the sign changes the same as for positive roots.

In your first problem there are 3 sign changes so there are 3 positive roots (one of which may be a repeated root). Changing the signs of the x^3 and x terms leads to no sign changes (all of the coefficients are negative) so there are no negative roots and the total is 3 roots, all positive.

In the 2'nd example, there are 6 sign changes so there are 6 positive roots. Changing the sign of the x^5, x^3, and x terms leaves no sign changes so there are no negative roots. Since a polynomial of order n has n roots, there are 8-6 = 2 imaginary or complex roots in the polynomial and 6 real roots (which may contain repeated roots)


Doug

Answer 2

Since you want direction, probably this example can help you understand...

The polynomial

X^3 + X^2 -X -1

has one sign change between the second and third terms. Therefore it has exactly 1 positive root.

Negating the odd-power terms gives

-X^3 +X^2+X -1

This polynomial has two sign changes, so the original polynomial has 2 or 0 negative roots.

The polynomial factors easily as

(X+1)^2 (X-1)

so the roots are -1 (twice) and 1.