<s>[INST] What is the remainder $ax+b$ when a cubic polynomial $P(x)$ is divided by $x^2-1$? If a cubic polynomial $P(x)$ with real coefficients has remainder 3 when divided by $x-1$ and remainder -7 when divided by $x+1$,
What is the remainder $ax+b$ when divided by $x^2-1$?
I see that since $$P(n) = ax^3 + bx^2 + cx + d$$ then $$P(1) = a + b + c + d = 3 $$ while $$P(-1) = -a + b - c + d = -7$$
I know $x^2-1$ is $(x-1)(x+1)$
I'm stuck here - how can I solve this? Could you please help? 
 [/INST] We have
\begin{align}
p(x) & = (x-1)q(x) +3\\
p(x) & = (x+1)r(x) - 7\\
p(x) & = (x^2-1)h(x) + (ax+b)
\end{align}
This gives us $p(1) = 3$, $p(-1) = -7$. Hence, we have
$$a+b = 3 \text{ and }-a+b = -7$$
Trust you can finish now.
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