Solve the equation

$$
x^{4}-2 \sqrt{3} x^{2}+x+3-\sqrt{3}=0
$$
Solution. The fourth-degree polynomial on the left side of the equation can be transformed into the product of two quadratic polynomials in several steps. The polynomial $x^{4}-2 \sqrt{3} x^{2}+3$ can be factored, as it is a perfect square. Thus, the equation becomes

$$
\left(x^{2}-\sqrt{3}\right)^{2}+x-\sqrt{3}=0
$$

Preparing to use the known identity $a^{2}-b^{2}=(a-b)(a+b)$, we subtract and add $x^{2}$ to the left side.

$$
\left(x^{2}-\sqrt{3}\right)^{2}-x^{2}+x^{2}+x-\sqrt{3}=0
$$

After applying the identity, we get

$$
\left(x^{2}-x-\sqrt{3}\right)\left(x^{2}+x-\sqrt{3}\right)+x^{2}+x-\sqrt{3}=0
$$

Now we can also factor out $\left(x^{2}+x-\sqrt{3}\right)$:

$$
\left(x^{2}+x-\sqrt{3}\right)\left(x^{2}-x+1-\sqrt{3}\right)=0 .
$$

A product is zero if and only if one of its factors is zero. First, consider the case when

$$
\left(x^{2}+x-\sqrt{3}\right)=0
$$

Using the quadratic formula, we get two solutions:

$$
x_{1,2}=\frac{-1 \pm \sqrt{1+4 \sqrt{3}}}{2}
$$

Now consider the other case:

$$
x^{2}-x+1-\sqrt{3}=0
$$

Applying the quadratic formula:

$$
x_{3,4}=\frac{1 \pm \sqrt{1-4+4 \sqrt{3}}}{2}=\frac{1 \pm \sqrt{4 \sqrt{3}-3}}{2}
$$

In the solution process, we used equivalent steps, and all four obtained real numbers are solutions to the equation.

Note: The factorization of the equation can be shortened with a non-standard idea. The equation

$$
x^{4}-2 \sqrt{3} x^{2}+x+3-\sqrt{3}=0
$$

can be viewed as a quadratic in $\sqrt{3}$.

$$
(\sqrt{3})^{2}-\left(2 x^{2}+1\right) \sqrt{3}+x^{4}+x=0 .
$$

Now let's write the quadratic formula:

$$
\sqrt{3}=\frac{2 x^{2}+1 \pm \sqrt{4 x^{4}+4 x^{2}+1-4 x^{4}-4 x}}{2}=\frac{2 x^{2}+1 \pm|2 x-1|}{2} .
$$

The two roots of the equation are

$$
\sqrt{3}=x^{2}+x, \text { and } \sqrt{3}=x^{2}-x+1
$$

From here, the factorized form of the equation, the product form, can be derived:

$$
\left(\sqrt{3}-x^{2}-x\right)\left(\sqrt{3}-x^{2}+x-1\right)=0 .
$$

From here, the solution can be completed as before.