Think about Question 1 Given $f(x)=4 x^{4}-2 x^{3}-16 x^{2}+5 x+9, g(x)=2 x^{3}-x^{2}-5 x+4$, find $u(x)$ and $v(x)$, such that $u(x) f(x)+v(x) g(x)=(f(x), g(x))$.
Solving using the Euclidean algorithm, we get
$$
\begin{array}{l}
f(x)=g(x) \cdot 2 x+\left(-6 x^{2}-3 x+9\right) \\
g(x)=\left(-6 x^{2}-3 x+9\right)\left(-\frac{1}{3} x+\frac{1}{3}\right)+(-x+1)
\end{array}
$$

Since $-6 x^{2}-3 x+9=(-x+1)(6 x+9)$.
Thus, $(f(x), g(x))=x-1$.
Rearranging equation (2) and substituting into equation (1), we have $(f(x), g(x))=x-1$
$$
\begin{array}{l}
=\left(-6 x^{2}-3 x+9\right)\left(-\frac{1}{3} x+\frac{1}{3}\right)-g(x) \\
=[f(x)-2 x g(x)]\left(-\frac{1}{3} x+\frac{1}{3}\right)-g(x) \\
=\left(-\frac{1}{3} x+\frac{1}{3}\right) f(x)+\left(\frac{2}{3} x^{2}-\frac{2}{3} x-1\right) g(x) .
\end{array}
$$

Therefore, we get $u(x)=-\frac{1}{3} x+\frac{1}{3}, v(x)=\frac{2}{3} x^{2}-\frac{2}{3} x-1$.