Is it possible to prove that $P(x)$ is unique?

Question

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?

Answer 1

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: $$\color{green}{\begin{cases} P(x)=g(x)x^2-x(g(x)-2)-2g(x)+2 \\ g(x)=x^2-x+1 \end{cases} \Longrightarrow P(x)=x^4-2x^3+3x}$$

Answer 2

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)$.