Is it possible to prove that $P(x)$ is unique? 
Let $P(x)$ be a polynomial.
If $P(x) \bmod (x+1)=0$ and $P(x) \bmod (x-2)=6$, then here are my questions:


i) What is the general form of the $P(x)$ ?
ii) What is the minimal degree of polynomial for $P(x)$ ?
iii) Is the polynomial $P(x)$ unique? Can we prove this uniqueness?

I have an example for $P(x)$.
$$P(x)=
6x^2-4x-10$$
Can we prove that, this polynomial is unique?
 A: hint
$$General \;\; form\;\; of \;\; P(x)$$
By hypothesis
$$P(x)=(x+1)Q(x)$$
$$P(x)=6+(x-2)R(x)$$
with $$d°(Q)=d°(R)=d°(P)-1$$
and
$$d°(P(x))\ge 1$$
$$Minimal \;\; degree \;\;and \;\;Uniqueness$$
Assume $P(x)=a(x+1)$.
then
$R(x)=b$ with
$$a(x+1)=6+b(x-2)=bx+6-2b$$
thus
$$a=b \text{ and } a=6-2b$$
So $$a=b=2$$
and
$$P(x)=2(x+1)=6+2(x-2)$$
The minimal degree is $ 1$ and the unique solution is
$$\boxed{P(x)=2x+2}$$
Remark.
The polynome $6x^2-4x-10$ you found is the case where  $ Q(x)=(3x-5)$.
A: My work:
$$\begin{cases} P(x)=k_1(x)(x+1) \\ P(x)=k_2(x)(x-2)+6 \end{cases} \Longrightarrow (x+1)k_1(x)=k_2(x)(x-2)+6 ​ \Longrightarrow (x+1)k_1(x)=k_2(x)(x+1)-3(k_2(x)-2) \Longrightarrow (k_2(x)-2) \mod (x+1)=0 \Longrightarrow k_2(x)=g(x)(x+1)+2 \\ \\ k_1(x)= \dfrac{k_2(x)(x-2)+6}{x+1} =\dfrac{(g(x)(x+1)+2) \times (x+1)+2)(x-2)+6}{x+1}=\dfrac{x(x+1)g(x)-2g(x)(x+1)+2x+2}{x+1}=\dfrac{(x+1)(xg(x)-2g(x)+2)}{x+1}=xg(x)-2g(x)+2=g(x)(x-2)+2 \\ P(x)=(x+1)(xg(x)-2g(x)+2)=g(x)x^2-x(g(x)-2)-2g(x)+2$$
$$\color{red}{\boxed{{P(x)=g(x)x^2-x(g(x)-2)-2g(x)+2}}}$$
here, $g(x)$ is an any polynomial.
$$ \color{blue}{\begin{cases} P(x)=g(x)x^2-x(g(x)-2)-2g(x)+2 \\ g(x)=0 \end{cases} \Longrightarrow P(x)=2x+2}$$
Random example:
$$\color{green}{\begin{cases} P(x)=g(x)x^2-x(g(x)-2)-2g(x)+2 \\ g(x)=x^2-x+1 \end{cases} \Longrightarrow P(x)=x^4-2x^3+3x}$$
