Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 341

\[\boxed{\mathbf{341}.}\]

\[x^{4} + 2x^{3} - 16x^{2} - 2x + 15 = 0\]

\[x_{1};x_{2};x_{3};x_{4} - корни\ \]

\[уравнения.\]

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

\[\ x_{3} = - y_{3};\ \ x_{4} = - y_{4}.\]

\[Подставим:\]

\[1)\ y_{1} + y_{2} + y_{3} + y_{4} = 2;\]

\[2)\ y_{1}y_{2} + y_{2}y_{3} + y_{3}y_{4} +\]

\[+ y_{2}y_{4} + y_{1}y_{4}y_{1}y_{3} = - 16;\]

\[3)\ y_{1}y_{2}y_{3} + y_{2}y_{3}y_{4} + y_{1}y_{2}y_{4} +\]

\[+ y_{1}y_{3}y_{4} = - 2;\]

\[4)\ y_{1}y_{2}y_{3}y_{4} = - x_{1}x_{2}x_{3} = 15.\]

\[Получаем\ уравнение:\]

\[x^{4} - 2x^{3} - 16x^{2} + 2x +\]

\[+ 15 = 0.\]