College Algebra, Chapter 4, 4.4, Section 4.4, Problem 68

Determine how many positive and how many negative real zeros the polynomial $P(x) = x^8 - x^5 + x^4 - x^3 + x^2 - x + 1$ can have, using the Descartes' Rule of signs. Then determine the possible total number of real zeros.

We first look at $P(x)$

$P(x) = + x^8 - x^5 + x^4 - x^3 + x^2 - x + 1$

$P(x)$ has six variations in sign, so it has either $6, 4, 2$ or positive roots.

Now,


$
\begin{equation}
\begin{aligned}

P(-x) =& (-x)^8 - (-x)^5 + (-x)^4 - (-x)^3 + (-x)^2 - (-x) + 1
\\
\\
P(-x) =& + x^8 + x^5 + x^4 + x^3 + x^2 + x + 1

\end{aligned}
\end{equation}
$


So $P(-x)$ has no variation in sign. Thus, $P(x)$ has negative roots, making a total of either $7, 5, 3$ or $1$ real zeros. Since is a zero but is neither positive nor negative.