College Algebra, Chapter 4, 4.4, Section 4.4, Problem 64

Determine how many positive and how many negative real zeros the polynomial $P(x) = 2x^3 - x^2 + 4x - 7$ can have, using the Descartes' Rule of signs. Then determine the possible total number of real zeros.

$P(x)$ has three variations in sign, so it has either three or one positive roots.

Now,


$
\begin{equation}
\begin{aligned}

P(-x) =& 2 (-x)^3 - (-x)^2 + 4(-x) - 7
\\
\\
P(-x) =& -2x^3 - x^2 - 4x - 7

\end{aligned}
\end{equation}
$


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