Example 2. Factor the polynomial $x^{4}-3 x^{2}+9$.

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Let $x=10$, then
$$
\begin{aligned}
x^{4}-3 x^{2}+9 & =10^{4}-3 \times 10^{2}+9 \\
& =9709 \\
& =7 \times 19 \times 73 .
\end{aligned}
$$

These three numbers are all prime. Substituting $x=10$ back, 7 is $x-3$, 19 is $x+9$ or $2x-1$. Dividing the original expression by $x-3$ does not result in an integer, so $x-3$ is not a factor of the original expression. Similarly, $x+9$ is also not a factor of the original expression. In $2x-1$, since the coefficient of $x$ is 2, and the coefficient of the highest degree term in the original expression does not contain the factor 2, $2x-1$ is also not a factor of the original expression. Therefore, the original expression has no linear factors. Multiplying 7 and 19 gives 133. Thus, the only possible quadratic factor of the original expression is $x^{2}+3x+3$. Dividing the original expression by $x^{2}+3x+3$ yields $x^{2}-3x+3$, so the original expression can be factored as:
$$
\begin{array}{l}
x^{4}-3x^{2}+9 \\
=\left(x^{2}+3x+3\right)\left(x^{2}-3x+3\right)
\end{array}
$$