Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 307

\[\boxed{\mathbf{307}.}\]

\[P(x) = x^{5} + ax^{3} + bx^{2} + c;\ \]

\[\ x = - 2:\]

\[P( - 2) = - 32 - 8a +\]

\[+ 4b + c = 0.\]

\[P(x) = \left( x^{2} - 1 \right) \cdot Q(x) - 3x + 3\]

\[x = \pm 1 - корни.\]

\[P( - 1) = - 1 - a + b + c =\]

\[= 0 \cdot Q(x) - 3 \cdot ( - 1) + 3\]

\[P(1) = 1 + a + b + c =\]

\[= 0 \cdot Q(x) - 3 \cdot 1 + 3\]

\[\left\{ \begin{matrix} - 8a + 4b + c = 32 \\ - a + b + c = 7\ \ \ \ \ \ \\ a + b + c = - 1\ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\text{\ \ }\left\{ \begin{matrix} 2b + 2c = 6\ \ |\ :2\ \ \ \\ a = - 1 - b - c\ \ \ \ \ \ \\ - 8a + 4b + c = 32 \\ \end{matrix} \right.\ \text{\ \ \ \ }\]

\[\left\{ \begin{matrix} b = 3 - c\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ a = - 4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 32 + 4 \cdot (3 - c) + c = 32 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} a = - 4\ \ \ \ \\ b = 3 - c \\ c = 4\ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\left\{ \begin{matrix} a = - 4 \\ b = - 1 \\ c = 1\ \ \ \\ \end{matrix} \right.\ \]

\[Ответ:a = - 4;\ \ b = - 1;\ \ c = 1.\]