2018/19 AHSMC Part 1 Problem 10

From $p(-x) = -p(x)$ we have $p(0) = -p(0)$ implying $p(0) = 0$, thus $x = 0$ is a root. Further, $p(1) = -p(-1) = 0$ and $p(-\sqrt{2}) = -p(\sqrt{2}) = 0$ implying $1$ and $-\sqrt{2}$ are also roots. Since $p(x)$ is of degree $5$ with leading coefficient $1$, there is a unique polynomial satisfying the given conditions, namely $$p(x) = x(x-1)(x+1)(x-\sqrt{2})(x+\sqrt{2}) = x(x^2-1)(x^2-2).$$ Thus, $p(2) = 2(2^2 - 1)(2^2 - 2) = 12$. The answer is (C).