Let $P(x)=ax^{2014}-bx^{2015}+1$ and $Q(x)=x^2-2x+1$ such that $Q(x)|P(x)$, find $a+b$ 
Let $P(x)=ax^{2014}-bx^{2015}+1$ and $Q(x)=x^2-2x+1$ be the polynomials where $a$ and $b$ are real numbers. If polynomial $P$ is divisible by $Q$, what is the value of $a+b$.

This is what I have tried so far: Since $Q(x)|P(x)$ we have $P(1)=0$, therefore $a-b+1=0$. Problem is because we cannot obtain system of equations, because polynomial $Q(x)$ has double root at $x=1$. From equation $a-b+1$ we cannot find $a+b$, so how to find out value of $a+b$ or $a^2-b^2$?

Note that $Q(x) = (x-1)^2$, which means that $(x-1) \mid P(x)$ and $(x-1) \mid P'(x)$. In other words $x=1$ is a zero of both $P(x)$ and $P'(x)$. So after all you're left to solve the system of linear equation:
$$\begin{cases} a-b+1 = 0 \\ 2014a - 2015b = 0 \end{cases}$$
It's easy to conclude that $a=-2015$ and $b=-2014$
