College Algebra, Chapter 4, 4.3, Section 4.3, Problem 58

Prove that $\displaystyle c = \frac{1}{3}$ and $c = -2$ are zeros of $P(x) = 3x^4 - x^3 - 21x^2 - 11x + 6$ and find all other zeros of $P(x)$.

If $\displaystyle P \left( \frac{1}{3} \right) = 0$, then $\displaystyle x - \frac{1}{3} = 0$, so $x - 3$ is a factor. Similarly, if $P(-2) = 0$ then $x + 2 = 0$, so $x + 2$ is a factor. Using synthetic division twice







We see that


$
\begin{equation}
\begin{aligned}

P(x) =& 3x^4 - x^3 - 21x^2 - 11x + 6
\\
\\
P(x) =& \left( x - \frac{1}{3} \right) (x - 3) (3x^2 - 6x - 9)
\\
\\
P(x) =& 3 \left( x - \frac{1}{3} \right) (x - 3) (x + 3)(x - 1)

\end{aligned}
\end{equation}
$