2. Find the product of all roots of the equation $x^{4}+4 x^{3}-2015 x^{2}-4038 x+2018=0$.
Solution. Transform the equation to the form: $\left(x^{2}+2 x\right)^{2}-2019\left(x^{2}+2 x\right)+2018=0$. Introduce the substitution $\boldsymbol{t}=\boldsymbol{x}^{2}+2 \boldsymbol{x}$, we get the equation: $t^{2}-2019 t+2018=0$, the roots of which are $t_{1}=1, t_{2}=2018$. Then $x^{2}+2 x-1=0$ or $x^{2}+2 x-2018=0$, from which we have $x_{1,2}=-1 \pm \sqrt{2}$, $x_{3,4}=-1 \pm \sqrt{2019}$. Their product is 2018.

Answer: 2018.