Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 312

\[\boxed{\mathbf{312}.}\]

\[P(x) = M_{1}(x)(x + 3) + 10\]

\[P(x) = M_{2}(x)(x + 5) + 14\]

\[P(x) = M_{3}(x)\left( x^{2} + 8x + 15 \right) +\]

\[+ ax + b\]

\[x = - 3:\]

\[P( - 3) = 0 \cdot M_{1}(x) + 10 = 10;\]

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

\[= - 3a + b.\]

\[x = - 5:\]

\[P( - 5) = 0 \cdot M_{2}(x) + 14 = 14;\]

\[P( - 5) = 0 \cdot M_{3}(x) - 5a + b =\]

\[= - 5a + b.\]

\[\left\{ \begin{matrix} - 3a + b = 10 \\ - 5a + b = 14 \\ \end{matrix} \right.\ ( - )\]

\[2a = - 4\]

\[a = - 2.\]

\[b = 10 + 3a = 10 - 6 = 4.\]

\[Ответ:4 - 2x.\]