Employing the elementary symmetric polynomials $\left(s_{1}=\alpha_{1}+\alpha_{2}+\alpha_{3}+\alpha_{4}=\right. -2, s_{2}=\alpha_{1} \alpha_{2}+\alpha_{1} \alpha_{3}+\alpha_{1} \alpha_{4}+\alpha_{2} \alpha_{3}+\alpha_{2} \alpha_{4}+\alpha_{3} \alpha_{4}=0, s_{3}=\alpha_{1} \alpha_{2} \alpha_{3}+\alpha_{2} \alpha_{3} \alpha_{4}+\alpha_{3} \alpha_{4} \alpha_{1}+\alpha_{4} \alpha_{1} \alpha_{2}=0$ and $s_{4}=\alpha_{1} \alpha_{2} \alpha_{3} \alpha_{4}=2$ we consider the polynomial $P(x)=\left(x-\left(\alpha_{1} \alpha_{2}+\alpha_{3} \alpha_{4}\right)\right)\left(x-\left(\alpha_{1} \alpha_{3}+\alpha_{2} \alpha_{4}\right)\right)\left(x-\left(\alpha_{1} \alpha_{4}+\alpha_{2} \alpha_{3}\right)\right)$. Because $P$ is symmetric with respect to $\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}$, we can express the coefficients of its expanded form in terms of the elementary symmetric polynomials. We compute $P(x) =x^{3}-s_{2} x^{2}+\left(s_{3} s_{1}-4 s_{4}\right) x+\left(-s_{3}^{2}-s_{4} s_{1}^{2}+s_{4} s_{2}\right) =x^{3}-8 x-8 =(x+2)\left(x^{2}-2 x-4\right)$. The roots of $P(x)$ are -2 and $1 \pm \sqrt{5}$, so the answer is \{1 \pm \sqrt{5},-2\}.