, 2014

Solution:

Since $P(P(1)) = P(P(2)) = 0$, but $P(1) \neq P(2)$, it follows that $P(1) = 1 + b + c$ and $P(2) = 4 + 2b + c$ are the distinct roots of the polynomial $P(x)$. Thus, $P(x)$ factors:
$$\begin{aligned} P(x) &= x^{2} + b x + c \\ &= (x - (1 + b + c))(x - (4 + 2b + c)) \\ &= x^{2} - (5 + 3b + 2c)x + (1 + b + c)(4 + 2b + c) \end{aligned}$$
It follows that $-(5 + 3b + 2c) = b$, and that $c = (1 + b + c)(4 + 2b + c)$. From the first equation, we find $2b + c = -\frac{5}{2}$. Plugging in $c = -\frac{5}{2} - 2b$ into the second equation yields
$$-\frac{5}{2} - 2b = (1 + (-\frac{5}{2}) - b)(4 + (-\frac{5}{2}))$$
Solving this equation yields $b = -\frac{1}{2}$, so $c = -\frac{5}{2} - 2b = -\frac{3}{2}$.

Therefore,
$$P(0) = 0^{2} + b \cdot 0 + c = c = -\frac{3}{2}.$$