<s>[INST] 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?
 [/INST] 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:
$$\color{green}{\begin{cases} P(x)=g(x)x^2-x(g(x)-2)-2g(x)+2 \\ g(x)=x^2-x+1 \end{cases} \Longrightarrow P(x)=x^4-2x^3+3x}$$
</s>