AndrewTom wrote: The polynomial $p(x)$ is of degree $9$ and $p(x)-1$ is exactly divisible by $(x-1)^{5}$. Given that $p(x) + 1$ is exactly divisible by $(x+1)^{5}$, find $p(x)$. $P(x)=1+(x-1)^5Q(x)$ with $Q(x)$ some polynomial of degree $4$ $P(x)+1=2+(x-1)^5Q(x)=(x+1)^5R(x)$ with $R(x)$ some polynomial of degree $4$ Setting there $x=-1$, we get $Q(-1)=\frac 1{16}$ Derivating $k$ times, with $k\in[1,4]$, we get $\sum_{i=0}^k\binom ki\frac{5!}{(5-i)!}(x-1)^{5-i}Q^{(k-i)}(x)$ $=\sum_{i=0}^k\binom ki\frac{5!}{(5-i)!}(x+1)^{5-i}R^{(k-i)}(x)$ Setting $x=-1$ in these four equalities, we get $\sum_{i=0}^k\binom ki\frac{5!}{(5-i)!}(-2)^{5-i}Q^{(k-i)}(-1)=0$ and so : $Q(-1)=\frac 1{16}$ $-32Q'(-1)+80Q(-1)=0$ $-32Q^{2'}(-1)+160Q'(-1)-160Q(-1)=0$ $-32Q^{3'}(-1)+240Q^{2'}(-1)-480Q'(-1)+240Q(-1)=0$ $-32Q^{4'}(-1)+320Q^{3'}(-1)-960Q^{2'}(-1)+960Q'(-1)-240Q(-1)=0$ $Q(-1)=\frac 1{16}$ $Q'(-1)=\frac 5{32}$ $Q^{2'}(-1)=\frac {15}{32}$ $Q^{3'}(-1)=\frac {105}{64}$ $Q^{4'}(-1)=\frac {210}{32}$ So $Q(x+1)=\frac{70}{256}x^4+\frac{35}{128}x^3+\frac{15}{64}x^2+\frac{5}{32}x+\frac 1{16}$ $Q(x)=\frac{35x^4+175x^3+345x^2+325x+128}{128}$ And $P(x)=1+\frac{(x-1)^5(35x^4+175x^3+345x^2+325x+128)}{128}$ $=\boxed{\frac{35x^9-180x^7+378x^5-420x^3+315x}{128}}$

It is easy to see we need $x^5|1+p(x-1)=f(x)(x-2)^5+2$ for some polynomial $f$ of degree 4. Obviously $x^5|1+p(x-1)$ when $f(x)=\frac{(x^5-2^5)^5}{2^{24}(x-2)^5}$, . To reduce degree of $f$ to 4, we only need to calculate remainder of $f$ divided by $x^5$. Or the Taylor expansion of $\frac{2}{(2-x)^5}$ up to degree 4, which is $\frac1{16}\sum_{k=0}^{4}C_k^{k+4}(\frac{x}2)^k$. So the answer is $p(x)=1+\frac1{16}\sum_{k=0}^{4}C_k^{k+4}(\frac{x+1}2)^k(x-1)^5$. If we generalize 5 to $n$ for a $p(x)$ of degree $2n-1$, we have $p(x)=1+\frac1{(-2)^{n-1}}\sum_{k=0}^{n-1}C_k^{k+n-1}(\frac{x+1}2)^k(x-1)^n$.