Expert: Paul Klarreich - 8/1/2009

Question
USE dESCRATES RULE TO GIVE THE POSSIBLE COMBINATIONS OF POSITIVE,NEGATIVE,AND IMAGINARYZEROS OF THE FOLLOWINGS:
1.P(X)=X^3+X^2-9X-9
2.P(X)=X^4+X^3-19X^2+x-20
3.P(X)=x^4-7x^3+5x^2-49x-30
4.P(X)=7x^5-4x^4-3x^3-x^2+x-3
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
7.P(X)=

Answer
Questioner: AMY
Country: Saudi Arabia
Category: Advanced Math
Private: Yes  << Changed.
Subject: hi help
Question: USE desCartes RULE TO GIVE THE POSSIBLE COMBINATIONS OF POSITIVE,NEGATIVE,AND IMAGINARYZEROS OF THE FOLLOWINGS:
1.P(X)=X^3+X^2-9X-9
2.P(X)=X^4+X^3-19X^2+x-20
3.P(X)=x^4-7x^3+5x^2-49x-30
4.P(X)=7x^5-4x^4-3x^3-x^2+x-3
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
7.P(X)=
.................................
Hi, Amy,

let's try ONE of them:

4.P(X) = 7x^5 - 4x^4 - 3x^3 - x^2+ x - 3

This has 5 zeroes. (Any polynomial has the same number as its degree.)
Descartes' rule 1:
The number of positive roots does not exceed the number of changes of sign in P(x).

There are 3 changes of sign, therefore at most 3 positive zeroes.

...................
Descartes' rule 2:
The number of negstive roots does not exceed the number of changes of sign in P(-x).


If we change odd-power terms:

P(-X) = -7x^5 - 4x^4 + 3x^3 - x^2 - x - 3
There are 2 changes of sign, therefore at most 2 negative zeroes.

Imaginary roots must occur in pairs, so the possibilities are:

Imag   Positive  Negative.
 4        1        0
 4        0        1
 2        3        0
 2        2        1
 2        1        2
 0        3        2

I think that does it.
You should be able to do the others yourself, now.