Polynomial Division!

Let's call $y = x^2$ and the polynomial $q(y) = P(x) \Rightarrow q(y) = y^8 + y^7 + y^6 + ... + y + 1 \rightarrow q(y) = (y - 2) r(y) + C$ where $r(y)$ is a polynomial in $y$ and $C$ is a constant, and $C$ is the same value defined above in the question $\Rightarrow q(2) = C$ ( remainder factor theorem ) and $q(2) = 2^{8} + 2^{7} + 2^{6} + .... + 2^1 + 1 \Rightarrow C$ has $\boxed{9}$ one's in base 2.

Note.- (Of course, I'm leting $P:\mathbb{R} \longrightarrow \mathbb{R}$ . If it was $P:\mathbb{Z}_{2} \longrightarrow \mathbb{Z}_{2}$ then the answer would be $C = 1 = 1_{2)}$ and $C$ would have 1 only one in base 2. )