Chapter 8.6, Problem 27WE

To determine

To find one real root of the equation by inspection and then show using Descartes’ rule that there are no other real roots.

Answer to Problem 27WE

Real root of the equation is 1 and there is no other real root.

Explanation of Solution

Given information:

The given equation is $x^5-x^4+2x^3-2x^2+3x-3=0$

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 :

Inspect if 1 is a root of equation,

  $\begin{array}{l}{x}^5-x^4+2x^3-2x^2+3x-3=0\\ 1^5-1^4+2\cdot 1^3-2\cdot 1^2+3\cdot 1-3=0\left[\text{ substitute }x=1\right]\\ 1-1+2-2+3-3=0\\ 0=0\end{array}$

  $x=1$ Satisfies the equation hence $\left(x-1\right)$ is a factor of the polynomial

Let $P\left(x\right)=x^5-x^4+2x^3-2x^2+3x-3$

Therefore $P\left(x\right)=\left(x-1\right)Q\left(x\right)$

Using long division, $Q\left(x\right)=P\left(x\right)÷\left(x-1\right)=x^4+2x^2+3$

The resulting polynomial is $P\left(x\right)=x^4+2x^2+3$

1. $P\left(x\right)$ Has no variation in sign, so by part (1) of Descartes’ rule the number of positive real roots of $P\left(x\right)=0$ is 0.
2. $P\left(-x\right)=x^4+2x^2+3$

$P\left(-x\right)$ has no variation in sign, so by part (2) of Descartes’ rule the number of negative real roots of $P\left(x\right)=0$ is 0.

Therefore it is proved that there is no other real roots.