Gọi  $ r_1,r_2,r_3 $  là số dư phép chia đa thức  $ Q(x) $  cho  $ x - a,x - b,x - c $  trong đó $3$ số $a, b, c$ đôi một khác nhau. Tìm đa thức dư  $ R(x) $ của phép chia đa thức  $ Q(x) $  cho tích  $ (x-a)(x-b)(x-c) $
Ta có: $ R\left( x \right) = m{x^2} + nx + p $
Theo giả thiết, ta suy ra m, n, p là nghiệm của hệ:
 $ \left\{ \begin{array}{l}
m{a^2} + na + p = {r_1}\\
m{b^2} + nb + p = {r_2}\\
m{c^2} + nc + p = {r_3}
\end{array} \right. $
Ta suy ra:
 $ \begin{array}{l}
m = \frac{{{r_1}}}{{\left( {a - b} \right)\left( {a - c} \right)}} + \frac{{{r_2}}}{{(b - c)(b - a)}} + \frac{{{r_3}}}{{\left( {c - a} \right)\left( {c - b} \right)}}\\
n = \frac{{ - \left( {b + c} \right){r_1}}}{{\left( {a - b} \right)\left( {a - c} \right)}} - \frac{{\left( {c + a} \right){r_2}}}{{(b - c)(b - a)}} - \frac{{\left( {a + b} \right){r_3}}}{{\left( {c - a} \right)\left( {c - b} \right)}}\\
p = \frac{{bc{r_1}}}{{\left( {a - b} \right)\left( {a - c} \right)}} + \frac{{c{\rm{a}}{{\rm{r}}_2}}}{{(b - c)(b - a)}} + \frac{{ab{r_3}}}{{\left( {c - a} \right)\left( {c - b} \right)}}\\
\Rightarrow R\left( x \right)
\end{array} $

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