4. Let $f(x)$ be a polynomial function of degree 2016 whose 2016 zeroes have a sum of $S$. Find the sum of the 2016 zeroes of $f(2 x-3)$ in terms of $S$.
Answer: $\frac{1}{2} S+3024$
Solution: Let $r_{1}, r_{2}, \cdots, r_{2016}$ be the zeroes of $f(x)$. We can then write $f$ as
$$
f(x)=c\left(x-r_{1}\right)\left(x-r_{2}\right) \cdots\left(x-r_{2016}\right) \text {, }
$$
where $\sum_{i=1}^{2016} r_{i}=S$. Thus
$$
f(2 x-3)=c\left(2 x-3-r_{1}\right)\left(2 x-3-r_{2}\right) \cdots\left(2 x-3-r_{2016}\right)
$$
which has zeroes
$$
\frac{r_{1}+3}{2}, \frac{r_{2}+3}{2}, \ldots, \frac{r_{2016}+3}{2}
$$

This means that the required sum is
$$
\sum_{i=1}^{2016} \frac{r_{i}+3}{2}=\frac{1}{2}\left[\sum_{i=1}^{2016} r_{i}+3(2016)\right]=\frac{1}{2} S+3024 .
$$