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By synthetic division, we get that $p(x) = (x - 2)(x^3 - x^2 - 20x + 50)$. By Rational Root Theorem, the possible roots of $x^3 - x^2 - 20x + 50$ are $\pm 1, \pm 2, \pm 5, \pm 10, \pm 25$, and $\pm 50$. Testing values, we get that $x = -5$ works, so by synthetic division, we get that $p(x) = (x - 2)(x + 5)(x^2 - 6x + 10)$.

By the quadratic formula, the roots of $x^2 - 6x + 10$ are $\frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 10}}{2 \cdot 1} = \frac{6 \pm 2i}{2} = 3 \pm i$. Thus, the roots of $p(x)$ are $\boxed{-5, 2, 3 + i, \text{and } 3 - i}$.