help

This probably isn't how this problem is intended to be done but it's all I can come up with.

$\text{Let }p(x) = x^{2017}+Ax+B\\ \text{We'll expand }p(x) \text{ as a Taylor series about }(x+1)\\ p(x) = \sum \limits_{k=0}^\infty~\dfrac{p^{(k)}(-1)(x+1)^k}{k!}\\ p(x) \text{ being divisible by }(x+1)^2 \text{ means that the first two terms must be 0}\\ p^{(0)}(-1) = p(-1) = -1 -A +B = 0\\ p^{(1)}(-1) = 2017(-1)^{2016} + A = 2017+A = 0\\ A=-2017\\ B = -2016$

$\text{Note: }p^{(k)}(x) = \dfrac{d^kp}{dx^k}(x)$