The polynomial $f(x)$ satisfies the equation $f(x)-f(x-2)=(2 x-1)^{2}$ for all $x$. If $p$ and $q$ are the coefficients of $x^{2}$ and $x$, respectively, in $f(x)$, then $p+q$ is equal to
(A) 0
(B) $\frac{5}{6}$
(C) $\frac{4}{3}$
(D) 1
(E) $\frac{2}{3}$
Since the difference between $f(x)$ and $f(x-2)$ has degree 2 , then the degree of $f(x)$ is at least 2. If the degree of $f(x)$ was equal to 2 , then the $x^{2}$ terms of $f(x)$ and $f(x-2)$ would cancel when we subtracted. (Try this.) Therefore, the degree of $f(x)$ must be at least 3.

So we try $f(x)=a x^{3}+p x^{2}+q x+r$.

Thus,

$$
\begin{aligned}
f(x-2) & =a(x-2)^{3}+p(x-2)^{2}+q(x-2)+r \\
& =a\left(x^{3}-6 x^{2}+12 x-8\right)+p\left(x^{2}-4 x+4\right)+q(x-2)+r \\
& =a x^{3}+(-6 a+p) x^{2}+(12 a-4 p+q) x+(-8 a+4 p-2 q+r)
\end{aligned}
$$

We are told that $f(x)-f(x-2)=(2 x-1)^{2}$, so $\left[a x^{3}+p x^{2}+q x+r\right]-\left[a x^{3}+(-6 a+p) x^{2}+(12 a-4 p+q) x+(-8 a+4 p-2 q+r)\right]=4 x^{2}-4 x+1$

$$
6 a x^{2}+(-12 a+4 p) x+(8 a-4 p+2 q)=4 x^{2}-4 x+1
$$

Comparing coefficients,

$$
\begin{aligned}
6 a & =4 \\
-12 a+4 p & =-4 \\
8 a-4 p+2 q & =1
\end{aligned}
$$

From the first equation, $a=\frac{2}{3}$.

Substituting $6 a=4$ into the second equation, we obtain $-8+4 p=-4$ or $p=1$.

Substituting these two values into the third equation, we obtain $8\left(\frac{2}{3}\right)-4(1)+2 q=1$ or $q=-\frac{1}{6}$.

Therefore, $p+q=1+\left(-\frac{1}{6}\right)=\frac{5}{6}$.

Answer: (B)