College Algebra, Chapter 4, Chapter Review, Section Review, Problem 38

If $P(x) = 6x^4 + 3x^3 + x^2 + 3x + 4$.

a.) List all possible rational zeros.

b.) Determine the possible number of positive and negative real zeros using Descartes' Rule of signs.



a.) The possible rational zeros are the factors of $4$ divided by the factors of $6$ which are $\displaystyle \pm 1, \pm 2, \pm 4, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}, \pm \frac{2}{3}$ and $\displaystyle \pm \frac{4}{3}$.

b.) Since the function doesn't change sign, then the possible number of positive real zeros is . And if


$
\begin{equation}
\begin{aligned}

P(-x) =& 6(-x)^4 + 3(-x)^3 + (-x)^2 + 3(-x) + 4
\\
\\
=& 6x^4 - 3x^3 + x^2 - 3x + 4

\end{aligned}
\end{equation}
$



Since $P(-x)$ has four variations in signs, then it shows that the function has either $4, 2$ or negative real zeros.