Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 313

\[\boxed{\mathbf{313}.}\]

\[P(x) = M_{1}(x)(x + 2) + 6;\]

\[P(x) = M_{2}(x)(x - 3) + 26;\]

\[P(x) = M_{3}(x)(x + 4) + 12;\]

\[P(x) =\]

\[= M_{4}(x)(x + 2)(x + 3)(x + 4) +\]

\[+ ax^{2} + bx + c.\]

\[x = - 2:\]

\[P( - 2) = 0 \cdot M_{1}(x) + 6 = 6;\]

\[P( - 2) = 0 \cdot M_{4}(x) + 4a - 2b +\]

\[+ c = 4a - 2b + c.\]

\[x = 3:\]

\[P(3) = 0 \cdot M_{2}(x) + 26 = 26;\]

\[P(3) = 0 \cdot M_{4}(x) + 9a + 3b +\]

\[+ c = 9a + 3b + c.\]

\[x = - 4:\]

\[P( - 4) = 0 \cdot M_{3}(x) + 12 = 12;\]

\[P( - 4) = 0 \cdot M_{4}(x) + 16a -\]

\[- 4b + c = 16a - 4b + c.\]

\[\left\{ \begin{matrix} 4a - b + c = 6 \rightarrow c = 6 - 4a + 2b \\ 9a + 3b + c = 26\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 16a - 4b + c = 12\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 9a + 3b + 6 - 4a + 2b = 26\ \ \\ 16a - 4b + 6 - 4a + 2b = 12 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 5a + 5b = 20\ \ |\ :5 \\ 12a - 2b = 6\ \ |\ :2 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} a + b = 4\ \ \ \\ 6a - b = 3 \\ \end{matrix} \right.\ ( + )\]

\[7a = 7\]

\[a = 1;\]

\[b = 4 - a = 4 - 1 = 3;\]

\[c = 6 - 4a + 2b =\]

\[= 6 - 4 + 6 = 8.\]

\[Ответ:x^{2} + 3x + 8.\]