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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?

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    $\begingroup$ You should look up Lagrange Interpolating polynomial. $$\begin{align} \text{remainder} &= 2\frac{(x+1)(x-2)}{(1+1)(1-2)} + 6\frac{(x-1)(x-2)}{(-1-1)(-1-2)} + 3\frac{(x-1)(x+1)}{(2-1)(2+1)}\\ &= -(x+1)(x-2) + (x-1)(x-2) + (x-1)(x+1)\\ &= x^2 - 2x + 3 \end{align} $$ $\endgroup$ – achille hui Jan 13 '15 at 19:17
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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 $a,b,c$

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  • $\begingroup$ We have not learned yet about Bezout's theorem, isn't there a simpler solution? $\endgroup$ – cristid9 Jan 13 '15 at 19:03
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    $\begingroup$ @cristid9 Take a look at my edit. $\endgroup$ – kingW3 Jan 13 '15 at 19:06
  • $\begingroup$ I have found the values of a, b and c, but I still don't know the value of $Q_{4}(x)$ and I can't find the polynomial without knowing the value of $Q_{4}(x)$. What should I do? $\endgroup$ – cristid9 Jan 13 '15 at 20:23
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    $\begingroup$ You don't need $Q_4(x)$ since you're looking only for the reminder. $\endgroup$ – kingW3 Jan 13 '15 at 20:26
  • $\begingroup$ Sorry, I am a little bit silly today. $\endgroup$ – cristid9 Jan 13 '15 at 20:29
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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 $P(x)=2p(x)+6q(x)+3r(x)$.

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Hint $\ f(x) = 2\! +\! (x\!-\!1)\left[a\! +\! (x\!-\!2)(b\!+\! (x\!+\!1)g(x))\right],$ $\,f(2) = 3\,\Rightarrow\, a=\_\_,\ f(-1) = 6\,\Rightarrow\, b = \_\_$

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