Chapter 8.6, Problem 19WE

To determine

To make the list of possibilities of nature of roots of equation.

Answer to Problem 19WE

Since $P\left(x\right)=0$ has five roots in all, there are four possibilities.

Number of Positive real roots	Number of negative real roots	Number of imaginary roots
3	0	2
3	2	0
1	0	4
1	2	2

Explanation of Solution

Given information:

The given equation is $x^5-x^3-x^2+x-2=0$

Formula used:

Descartes’ Rule of Signs

Let $P\left(x\right)$ be a simplified polynomial with real coefficients and terms arranged in decreasing degree of x.

1. The number of positive real roots of $P\left(x\right)=0$ equals to the number of variations of sign of $P\left(x\right)$ or is fewer than this number by an even integer.
2. The number of negative real roots of $P\left(x\right)=0$ equals to the number of variations of sign of $P\left(-x\right)$ or is fewer than this number by an even integer.

Calculation :

First write the polynomial

  $P\left(x\right)=x^5-x^3-x^2+x-2$

1. $P\left(x\right)$ Has only 3 variations in sign, so by part (1) of Descartes’ rule the number of positive real roots of $P\left(x\right)=0$ is 3 or 1.
2. $P\left(-x\right)=-x^5+x^3-x^2-x-2=0$

     $P\left(-x\right)$ Has two variations in sign, so by part (2) of Descartes’ rule the number of negative real roots of $P\left(x\right)=0$ is 2 or 0.

3. Since $P\left(x\right)=0$ has five roots in all, there are four possibilities.

Number of Positive real roots	Number of negative real roots	Number of imaginary roots
3	0	2
3	2	0
1	0	4
1	2	2