College Algebra, Chapter 4, 4.3, Section 4.3, Problem 56

If $P(x) = x^4 + 3x^3 - 16x^2 - 27x + 63$. Use the factor theorem to show that $x - c$ is a factor of $P(x)$ for $c = 3$ and $c = -3$.

If $P(3) = 0$, then $x - 3 = 0$, so $x - 3$ is a factor and if $P(-3) = 0$, then $x + 3 = 0$, so $x + 3$ is a factor. So using synthetic division twice







We see that


$
\begin{equation}
\begin{aligned}

P(x) =& x^4 + 3x^3 - 16x^2 - 27x + 63
\\
\\
P(x) =& (x - 3)(x + 3)(x^2 + 3x - 7)
\\
\\
\text{ or } &
\\
\\
P(x) =& (x -3)(x + 3) \left( x - \frac{3 + \sqrt{37}}{2} \right) \left( x - \frac{3 - \sqrt{37}}{2} \right)

\end{aligned}
\end{equation}
$