<s>[INST] Finding $ a $ where $ P(x)-a $ divides with $ (x+1)^4 $ and $ P(x)+a $ divides with $ (x-1)^4 $ Let $ P(x) $ be a $ 7 $ degree polynomial with the coefficient of $ x^7 $ equal to $ 1 $. Let $ a \in\mathbb{R} $ such that $ P(x)-a $ divides with $ (x+1)^4 $ and  $ P(x)+a $ divides with $ (x-1)^4 $
$ 1 ) $ Find the coefficient of $ x^5 $
The answer should be $ -\frac{21}{5} $
$ 2 ) $ Find $ a $
The answer should be $ \frac{16}{5} $
 [/INST] $(x+1)^4 \mid (P(x)-a) \implies P(x) = f(x)(x+1)^4+a \implies (x+1)^3 \mid P'(x)$.  
Similarly we have $(x-1)^3 \mid P'(x)$, so putting these together with $P(x)$ being monic of seventh degree, $P'(x) = 7(x^2-1)^3$.
$\implies P(x) = x^7 -\frac{21}5x^5+7x^3-7x + C$, 
Now $P(1) = -a, P(-1) = a \implies C-\frac{16}5=-a, C + \frac{16}5=a \implies C=0, a = \frac{16}5$.
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