Expert: Sherry Wallin - 8/2/2009

Question
USE dESCRATES RULE TO GIVE THE POSSIBLE COMBINATIONS OF POSITIVE,NEGATIVE,AND IMAGINARYZEROS OF THE FOLLOWINGS
5.P(X)=x^4+2x^3-x^2+7x-5
6.P(X)=2x^4+x^3-3x^2-x+5
7.P(x)=3x^3+4x^2-7x-2


Answer
Amy~
    The first thing you need to do is understand the definition of Descartes Rule:
The rule states that if the terms of a single-variable polynomial with real coefficients are ordered by descending variable exponent, then the number of positive roots of the polynomial is either equal to the number of sign differences between consecutive nonzero coefficients, or less than it by a multiple of 2. Multiple roots of the same value are counted separately. As a corollary of the rule, the number of negative roots is the number of sign changes after negating the coefficients of odd-power terms (otherwise seen as substituting the negation of the variable for the variable itself), or less than it by a multiple of 2. (Compliments of Wikipedia)

So using the definition above I will explain the answer to your (5) above
5.P(X)=x^4+2x^3-x^2+7x-5

The problem is in descending order already and all coefficients are real numbers so the hypothesis is satisfied.

The number of positive roots is either equal to the number of sign differences (here there are 3) or less than it by a multiple of 2. Since this is a 4th degree polynomial there could only be 4 or less real roots anyway so this means there are either 1 or 3 positive roots.

By the corollary the number of negative roots is the number of sign changes after negating the coefficients of the odd-power terms or less than it by a multiple of 2. Here you would have:
x^4-2x^3-x^2-7x-5 by negating the cube power term and the 1st power term. There is only one sign change from +x^4 to -x^3 so that means that there is one negative root (zero). Given that that leaves 3 other roots and since complex roots come in conjugate pairs there can be at most 1 pair of imaginary zeros. From the positive root statement the number of positive zeros is 1 or 3 and knowing there are 4 roots tells us that there is 1 negative root and 1 positive root or 1 negative and 3 positive roots. If the latter is the case then there are no imaginary zeros and if the former is the case then there is 1 pair of imaginary zeros. You can use the rational zeros theorem to further determine if there is an imaginary pair. The rational roots theorem says if you look at all the factors of the constant and put those factors in the numerator and use all the factors of the leading coefficient as possible denominators that those are the only choices for rational roots like this: 5 is the constant and the only factors of 5 are +-1 and +-5. Since the leading coefficient is 1, the only factors of 1 are +-1, so the possible rational roots are:
-5/1, 5/1, 1/1, -1/1, -5/-1, 5/-1, 1/-1,-1/-1 and there are duplicates so

possible zeros: +-5,+-1. You can use synthetic division or just evaluate these possible zeros:
f(1) = 1+2-1+7-5 = 4, not a zero
f(-1) = 1 -2 -1 -7 -5 = -14 not a zero
f(1/5 = .2) = 1/625 + 2/125-1/25 +7/5 -5 = (1 + 10 -25 + 875 -3250)/625 !=0 not a zero
f(-1/5) = 1/625 -2/125- 1/25-7/5 - 5 = (1 - 10 - 25 - 875 - 3250)/625 != 0
Doing all this hasn't given us any new usable information just that the zeros aren't rational so the conclusion remains the same: there is either 1 pair of complex conjugate zeros or none so there is either 2 real or 4 real zeros and 1 of them is negative and there is either 1 or 3 positive roots.

Math Prof