Chapter 8.6, Problem 15OE

To determine

To write the possibilities of numbers of positive and negative real roots using Descartes’ rule of signs.

Answer to Problem 15OE

Since $P\left(x\right)=0$ has six roots in all, there are only one possibility.

Number of Positive real roots	Number of negative real roots
1	1

Explanation of Solution

Given information:

The given equation is $x^6-10=0$

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

  $P\left(x\right)=x^6-10$

1. $P\left(x\right)$ Has only one variation in sign, so by part (1) of Descartes’ rule the number of positive real roots of $P\left(x\right)=0$ is 1.
2. $P\left(-x\right)=x^6-10$

     $P\left(-x\right)$ Has only one variation in sign, so by part (2) of Descartes’ rule the number of negative real roots of $P\left(x\right)=0$ is 1.

3. Since $P\left(x\right)=0$ has six roots in all, there is only one possibility.

Number of Positive real roots	Number of negative real roots
1	1