Example 10 has the equation
$$
x^{4}-6 x^{3}+8 x^{2}+8 x-16=0
$$

has a root of $1+\sqrt{5}$. Solve this equation.
Solution: From the fact that irrational roots of a rational coefficient equation come in conjugate pairs, we know that $1-\sqrt{5}$ is also a root of the original equation. Therefore, the 4th-degree polynomial on the left side of the original equation contains the factor
$$
(x-1+\sqrt{5})(x-1-\sqrt{5})=x^{2}-2 x-4 \text {. }
$$

Considering the coefficient of the $x^{4}$ term is $1=1 \times 1$, and the constant term $-16=-4 \times 4$, we can assume
$$
\begin{array}{l}
x^{4}-6 x^{3} \div 8 x^{2}+8 x-16 \\
=\left(x^{2}-2 x-4\right)\left(x^{2}+a x+4\right) .
\end{array}
$$

Taking $x=1$, we get $-5=-5(1+a+4)$, which gives $a=-4$.
$\therefore$ The original equation can be transformed into
$$
\left(x^{2}-2 x-4\right)\left(x^{2}-4 x+4\right)=0 \text {. }
$$

Solving this, we get $x_{1,2}=1 \pm \sqrt{5}, x_{3,4}=2$.