[Example 5.2.2] Solve the equation:
$$
x^{4}-x^{2}+8 x-16=0 .
$$
For this equation, it can be solved by factoring, transforming the equation into
$$
\begin{array}{c}
\left(x^{2}+4\right)\left(x^{2}-4\right)+8 x-x^{2}=0 . \\
\left(x^{2}+4-x\right)\left(x^{2}-4+x\right)=0 .
\end{array}
$$

When $x^{2}+4-x=0$, the solutions to the equation are
$$
x_{1,2}=\frac{1 \pm \sqrt{15} \mathrm{i}}{2} .
$$

When $x^{2}-4+x=0$, the solutions to the equation are
$$
x_{3,4}=\frac{-1 \pm \sqrt{17}}{2} .
$$

This solution seems straightforward, but factoring the left side of the equation $x^{4}-x^{2}+8 x-16$ is not easy to think of, so this method is more challenging for those who are not proficient in factoring techniques.

Similar to Example 5.2.1, we can use the method of "reversing the roles," transforming a quartic equation into a quadratic equation. We temporarily treat $x$ as a known number and the known number 16 as an unknown.
Let $4=y$, then the given equation becomes
$$
\begin{array}{c}
y^{2}-2 x y+x^{2}-x^{4}=0 . \\
\Delta=4 x^{2}-4 x^{2}+4 x^{4}=4 x^{4} . \\
y=\frac{2 x \pm 2 x^{2}}{2}=x \pm x^{2} .
\end{array}
$$

Thus,
$$
4=x \pm x^{2} \text {. }
$$

That is,
$$
x^{2}+x-4=0, x^{2}-x+4=0 .
$$

This yields the four solutions to the equation:
$$
x_{1,2}=\frac{-1 \pm \sqrt{17}}{2}, x_{3,4}=\frac{1 \pm \sqrt{15} \mathrm{i}}{2} .
$$