4. [5 points] Solve the equation $3 x^{4}+x^{2}-8 x-4 x^{2}|x-4|+16=0$.

翻译结果：
4. [5 points] Solve the equation $3 x^{4}+x^{2}-8 x-4 x^{2}|x-4|+16=0$.
Answer: $-\frac{4}{3} ; 1 ; \frac{-1 \pm \sqrt{17}}{2}$.

Solution. The equation can be rewritten as $3 x^{4}+(x-4)^{2}-4 x^{2}|x-4|=0$ or $|x-4|^{2}-4 x^{2} \mid x-$ $4 \mid+3 x^{4}=0$. To factorize the left side, note that it represents a quadratic trinomial in terms of $y=|x-4|$ with a discriminant equal to $\left(4 x^{2}\right)^{2}-4 \cdot 3 x^{4}=$ $4 x^{4}$. Therefore, the roots $y_{1,2}$ are $\frac{4 x^{2}+2 x^{2}}{2}$, i.e., $y_{1}=3 x^{2}$ and $y_{2}=x^{2}$, and the equation becomes $\left(|x-4|-3 x^{2}\right)\left(|x-4|-x^{2}\right)=0$.

The product is zero if and only if one of the factors is zero. From this, we get

$$
\left[\begin{array}{l}
x^{2}=x-4, \\
x^{2}=-x+4, \\
3 x^{2}+x-4=0, \\
3 x^{2}-x+4=0
\end{array} \Leftrightarrow \left[\begin{array}{l}
x^{2}+x-4=0 \\
3 x^{2}+x-4=0
\end{array} \Leftrightarrow \left[\begin{array}{l}
x=\frac{-1 \pm \sqrt{17}}{2} \\
x=-\frac{4}{3} \\
x=1
\end{array}\right.\right.\right.
$$