Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 387

\[\boxed{\mathbf{387}.}\]

\[ax^{3} - 2x^{2} - 5x + b = 0;\ \ \]

\[x_{1} = 1;\ \ x_{2} = - 2\]

\[P(1) = a - 2 - 5 + b = 0;\]

\[P( - 2) = - 8a - 8 + 10 +\]

\[+ b = - 8a + b + 2 = 0\]

\[\left\{ \begin{matrix} a + b = 7\ \ \ \ \ \ \ \ \ \\ - 8a + b = - 2 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} a = 7 - b\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ - 8 \cdot (7 - b) + b = - 2 \\ \end{matrix} \right.\ \]

\[- 56 + 8b + b = - 2\]

\[9b = 54\]

\[b = 6.\]

\[a = 7 - 6 = 1.\]

\[Получаем:\]

\[x^{3} - 2x^{2} - 5x + 6 = 0\]

\[P(x) = (x - 1)(x - 3)(x + 2).\]

\[Ответ:\ \ a = 1;\ \ b = 6;\ \ x = 3.\]