2. In the field of real numbers, solve the equation

$$
x^{3}+(3 \sqrt{2}-2) x^{2}-(1+\sqrt{2}) x-14(\sqrt{2}-1)=0,
$$

knowing that it has at least one integer root. Write any irrational roots in a simple form without radicals of irrational numbers.
SOLUTION. Let's rewrite the given equation in the form

$$
\sqrt{2}\left(3 x^{2}-x-14\right)+\left(x^{3}-2 x^{2}-x+14\right)=0 .
$$

From this, it follows that every integer root of the given equation must be a root of the equation

$$
3 x^{2}-x-14=0,
$$

otherwise, the irrational number $\sqrt{2}$ could be expressed from equation (1) as the ratio of two integers. This quadratic equation has roots $-2$ and $\frac{7}{3}$, of which only the first is a root of the original equation, as we can easily verify by substituting both numbers into the modified equation (1).

Thus, we have found the root $x_{1}=-2$ of the given cubic equation. The remaining roots are the roots of the quadratic equation obtained by dividing the original equation by the root factor $x+2$. We get

$$
\left(x^{3}+(3 \sqrt{2}-2) x^{2}-(1+\sqrt{2}) x-14(\sqrt{2}-1)\right):(x+2)=x^{2}+(3 \sqrt{2}-4) x-7(\sqrt{2}-1)=0 .
$$

The discriminant $D$ of the found quadratic equation is a positive irrational number

$$
D=(3 \sqrt{2}-4)^{2}+28(\sqrt{2}-1)=6+4 \sqrt{2} .
$$

To avoid square roots of irrational numbers in the notation of the remaining roots, let's seek the value of $\sqrt{D}$ in the form

$$
\sqrt{D}=\sqrt{6+4 \sqrt{2}}=a+b \sqrt{2}
$$

with rational coefficients $a, b$. These can be easily guessed, as upon squaring we get

$$
6+4 \sqrt{2}=a^{2}+2 b^{2}+2 a b \sqrt{2}
$$

from which $a^{2}+2 b^{2}=6$ and $2 a b=4$, so it is clear that the values $a=2$ and $b=1$ satisfy these equations. (Instead of guessing, we can substitute $b=2 / a$ and solve the quadratic equation $a^{2}+2 \cdot(2 / a)^{2}=6$ for the unknown $a^{2}$, which yields $a^{2}=2$ or $a^{2}=4$.)

Thus, $\sqrt{D}=2+\sqrt{2}$, so the remaining roots $x_{2,3}$ of the given equation are the numbers

$$
x_{2}=\frac{4-3 \sqrt{2}+(2+\sqrt{2})}{2}=3-\sqrt{2} \quad \text { and } \quad x_{3}=\frac{4-3 \sqrt{2}-(2+\sqrt{2})}{2}=1-2 \sqrt{2} .
$$

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