Find polynomial $f(x)$ based on divisibility properties of $f(x)+1$ and $f(x) - 1$ $f(x)$ is a fifth degree polynomial. It is given that $f(x)+1$ is divisible by $(x-1)^3$ and $f(x)-1$ is divisible by $(x+1)^3$. Find $f(x)$.
 A: We can clearly see that: $f(1)+1=0$ and $f(-1)-1=0$
We can write $f(x)-1=p(x)(x+1)^3$ and $f(x)+1=q(x)(x-1)^3$
By differentiation and double differentiation, you can see that
$f'(1)=0$ and $f''(1)=0$
AND 
$f'(-1)=0$ and $f''(-1)=0$
You got six conditions and six unknowns! 
[assume $f(x) = x^6+a_1x^5+a_2x^4+a_3x^3+a_4x^2+a_5x+a_6$]
A: If $(x-1)^3$ divides $f(x)+1$, then $(x-1)^2$ divides $f'(x)$.
If $(x+1)^3$ divides $f(x)-1$, then $(x+1)^2$ divides $f'(x)$.
As deg$\,f=5$, then deg$\,f'=4$, and hence $f'(x)=a(x-1)^2(x+1)^2=a(x^4-2x^2+1)$,
for some $a\in\mathbb R$.
Thus 
$$
f(x)=\frac{a}{5}x^5-\frac{2a}{3}x^3+ax+b,
$$
for some $b\in\mathbb R$.
Now, as $(x-1)^3$ divides $g(x)=f(x)+1$, in particular $f(1)+1=g(1)=0$. Thus
$$
-1=f(1)=\frac{a}{5}-\frac{2a}{3}+a+b. \tag{1}
$$ 
Similarly,
as $(x+1)^3$ divides $h(x)=f(x)-1$, in particular $f(-1)-1=g(-1)=0$. Thus
$$
1=f(-1)=-\frac{a}{5}+\frac{2a}{3}-a+b. \tag{2}
$$
Adding $(1)$ and $(2)$ we obtain that $b=0$, and thus $a=-15/8$.
Hence
$$
f(x)=-\frac{3}{8}x^5+\frac{5}{4}x^3-\frac{15}{8}x.
$$ 
A: Note that $(x-1)^3$ divides $f(x)+1$ and $f(-x)-1$, so $(x-1)^3$ divides the sum $f(x)+f(-x)$.
Similarly, we can note that $(x+1)^3$ divides $f(x)+f(-x)$.
Therefore, $(x+1)^3(x-1)^3$ divides $f(x)+f(-x)$, which has degree at most $5$.
This implies that $f(x)$ is an odd function.
We get $f(x)+1=(x-1)^3(ax^2+bx-1)$, (note that $f(0)$ must be $0$) and by comparing the degree $2$ and $4$ coefficients (which should be $0$), we get $a=-\frac{3}{8}$ and $b=-\frac{9}{8}$, which gives the answer of $$f(x)=-\frac{3}{8}x^5+\frac{5}{4}x^3-\frac{15}{8}x$$
A: Hint: Use the fact that $f^{\prime}(x)$ is divisible by $(x-1)^2$ and $(x+1)^2$.
A: Another ad hoc solution is the following:
$$
32 = \Big((x+1)-(x-1)\Big)^5 \\
= (x+1)^5 -5(x+1)^4(x-1) +10(x+1)^3(x-1)^2 \qquad \\ \qquad -10(x+1)^2(x-1)^3 +5(x+1)(x-1)^4
-(x-1)^5 \\
= (x+1)^3\Big((x+1)^2-5(x+1)(x-1)+10(x-1)^2\Big) \qquad \\ \qquad 
-(x-1)^3\Big(10(x+1)^2-5(x+1)(x-1)+(x-1)^2\Big)
$$
so
$$
f(x) =
-\frac1{16} (x+1)^3\Big((x+1)^2-5(x+1)(x-1)+10(x-1)^2\Big) +1 =
-\frac1{16} (x-1)^3\Big(10(x+1)^2-5(x+1)(x-1)+(x-1)^2\Big) -1
$$
is $a$ solution.

For solving the general problem, we can repeat the usual proof of the Chinese Remainder Theorem. To solve the system
$$
f(x) \equiv r_1(x) \pmod{m_1(x)} \\
f(x) \equiv r_2(x) \pmod{m_2(x)}
$$
where $m_1$ and $m_2$ are co-prime, using Euclid's algorithm we construct some polynomials $p(x)$ and $q(x)$ such that 
$$ p(x)\cdot m_1(x) + q(x)\cdot m_2(x) = 1. $$
Then the solution will be
$$
f = (q \cdot m_2 \cdot r_1 + p \cdot m_1 \cdot r_2) \mod m_1m_2.
$$
A: f(x)+1 = (x-1)³*r(x) ⇒ f '(x)=(x-1)²*r₁(x) ... (1)
f(x)-1 = (x+1)³*s(x) ⇒ f '(x)=(x+1)²s₁(x) ... (2)
From (1) and (2) ⇒ f '(x) has double root 1 and double root -1 ⇒ 
f '(x)=k(x-1)²(x+1)² = k*(x²-1)² = k*(x⁴-2x²+1), k∊R.
Then f(x)= ∫(kf '(x))dx = k∫(x⁴-2x²+1)dx = k*(x⁵/5 - 2x³/3 + x + C)
Because 
f(x)+1 = (x-1)³r(x) ⇒ f(1)+1=0 ⇒ k*(8/15+C)+1=0 ... (3)
f(x)-1 = (x+1)³s(x) ⇒ f(-1)-1=0 ⇒ k(-8/15+C)-1=0 ... (4)
From (3) and (4) ⇒ C=0, k = -15/8
Then we have the answer: 
f(x) = (-15/8)*(x⁵/5 - 2x³/3 + x )
f(x) = -15x⁵/40 + 30x³/24 - 15x/8
