College Algebra, Chapter 4, 4.4, Section 4.4, Problem 66

Determine how many positive and how many negative real zeros the polynomial $P(x) = x^4 + x^3 + x^2 + x + 12$ can have, using the Descartes' Rule of signs. Then determine the possible total number of real zeros.

$P(x)$ has no variation in sign, so there are no positive roots.

Now,


$
\begin{equation}
\begin{aligned}

P(-x) =& (-x)^4 + (-x)^3 + (-x)^2 + (-x) + 12
\\
\\
P(-x) =& + x^4 - x^3 + x^2 - x + 2

\end{aligned}
\end{equation}
$


So $P(-x)$ has four variations in sign. Thus, $P(x)$ has either $4, 2$ or negative roots, making a total of either $5, 3$ or $1$ real zeros. Since is a zero but is neither positive nor negative.