Find a polynomial given the remainders of the division of that polynomial with 3 other polynomials 
A polynomial from $ \mathbb{C}[x]$ divided by $ x - 1$, $x + 1$, $ x -2$ has the remainders 2, 6 and 3. Find the remainder of the division of that polynomial by $(x-1)(x+1)(x-2)$

The degree of the expression $(x-1)(x+1)(x-2)$ is 3, so the degree of the polynomial that I am looking for is 4. So the polynomial should be of the form: $ax^4 + bx^3 +cx^2 + dx + e$. I have tried to use Horner's scheme and I got the following expressions: $a+b+c+d+e=2$, $a+c-b-d+e=6$ and $16a+8b+4c+2d+c=3$. But this is not enough in order to find the polynomial. Is this the right approach?
 A: Lets call this polynomial $P(x)$ than by the conditions you have that the polynomial $P(x)$ can be written as $$P(x)=(x-1)Q_1(x)+2\\P(x)=(x+1)Q_2(x)+6\\P(x)=(x-2)Q_3(x)+3$$ From this you can see that $P(1)=2,P(-1)=6,P(2)=3$
Now you can also write your polynomial as
$$P(x)=(x-1)(x+1)(x-2)Q_4(x)+ax^2+bx+c$$
Where $ax^2+bx+c$ is reminder when $P(x)$ is divided by $(x-1)(x+1)(x-2)$ plugging in $x=1,-1,2$ you can find $a,b,c$
A: It's a problem about Lagrange's interpolation polynomials. The hypothesis means that $P(1)=2$, $P(-1)=6$, P(2)=3$.
The simplest method  uses concepts from linear algebra. You first solve 3 easier problems: find polynomials $p, q, r$ such that
\begin{alignat*}{3}
&p(1)=1&&q(1)=0&&r(1)=0 \\
&p(-1)=0&\qquad&q(-1)=1&\qquad&r(-1)=0\\
&p(2)=0&&q(2)=0&&r(2)=1
\end{alignat*}
For instance,$\,\,p(x)=\dfrac{(x+1)(x-2)}{(1+1)(1-2)}= -\dfrac12(x^2-x-2)$.
Once you've computed $p(x), q(x), r(x)$, just take $P(x)=2p(x)+6q(x)+3r(x)$.
A: Hint $\ f(x) = 2\! +\! (x\!-\!1)\left[a\! +\! (x\!-\!2)(b\!+\! (x\!+\!1)g(x))\right],$  $\,f(2) = 3\,\Rightarrow\, a=\_\_,\ f(-1) = 6\,\Rightarrow\, b = \_\_$
