Harvard-MIT Mathematics Tournament

Solution:

Answer: $r > 3$, $r < \frac{49}{27}$. Because such roots to polynomial equations come in conjugate pairs, we seek the values $r$ such that $P(x) = r$ has just one real root $x$. Considering the shape of a cubic, we are interested in the boundary values $r$ such that $P(x) - r$ has a repeated zero. Thus, we write
$$P(x) - r = x^{3} + x^{2} - x + (2 - r) = (x - p)^{2}(x - q) = x^{3} - (2p + q)x^{2} + p(p + 2q)x - p^{2}q$$
Then $q = -2p - 1$ and $1 = p(p + 2q) = p(-3p - 2)$ so that $p = 1/3$ or $p = -1$. It follows that the graph of $P(x)$ is horizontal at $x = 1/3$ (a maximum) and $x = -1$ (a minimum), so the desired values $r$ are $r > P(-1) = 3$ and $r < P(1/3) = 1/27 + 1/9 - 1/3 + 2 = 49/27$.