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> The quadratic polynomial $P(x)$ has a zero at $x=2$. The polynomial $P(P(x))$ has only one real zero at $x=5.$ Compute $P(0).$
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If we have that $P(x) = ax^2 +bx+ c$, we get from the first condition that $P(x) = (x-2)(bx+c).$

From here $P(P(x)) = (ax+bx+c -2)(b(ax+bx+c)+c)$, but this just looks very messy and doesn't seem to be helpful at all. Is there some other trick here I'm missing?