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 $b=-2014$
Since $Q$ divides $P$, we have that $x-1$ divides $P$ twice. Hence $1$ is a root of $P$ and of its derivative $P'(x)=2014ax^{2013}$. So $a-b^{2015}+1=0$ and $2014a=0$. Now we can solve to get $a=0$, but in that case $Q$ cannot divide $P$, since $P$ has degree $0$.
If we actually have $P(x)=ax^{2015}-bx^{2014}+1$ then its derivative is $2015ax^{2014}-2014bx^{2013}$ and we get instead the system $a-b+1=0, \, 2015a-2014b=0.$
