A vague problem

Note that the Taylor series for any function (in this case is a polynomial $P(x)$ ) centered at $x=2016$ is given by $\begin{aligned} P(x)=\frac{(x-2016)^0}{0!}P(2016)+\frac{(x-2016)^1}{1!}P'(2016)+\frac{(x-2016)^2}{2!}P''(2016)+\frac{(x-2016)^3}{3!}P'''(2016)+\cdots \end{aligned}$ Thus, dividing both sides with $(x-2016)^3$ will give us $\begin{aligned} \frac{P(x)}{(x-2016)^3}=\frac{(x-2016)^0}{0!(x-2016)^3}P(2016)+\frac{(x-2016)^1}{1!(x-2016)^3}P'(2016)+\frac{(x-2016)^2}{2!(x-2016)^3}P''(2016)+\frac{(x-2016)^3}{3!(x-2016)^3}P'''(2016)+\cdots \end{aligned}$ Since we only need to find the remainder, we can neglect all the terms with their numerators being a multiple of $(x-2016)^k$ where $k$ is a positive integer larger than or equal to 3. Therefore, the remainder is simply $P(2016)+P'(2016)(x-2016)+\frac{1}{2}P''(2016)(x-2016)^2$ .