Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 395

\[\boxed{\mathbf{395}.}\]

\[P(x) = x^{10} + bx^{9} + cx^{8}\]

\[P(x) =\]

\[= Q(x)\left( x + a_{1} \right)\left( x + a_{2} \right) = 0;\ \ \ \]

\[a_{1} \neq 0;\ \ a_{2} \neq 0.\]

\[P\left( - a_{1} \right) = a_{1}^{10} - b \cdot a_{1}^{9} +\]

\[+ c \cdot a_{1}^{8} = 0.\]

\[P\left( - a_{2} \right) = a_{2}^{10} - b \cdot a_{2}^{9} +\]

\[+ c \cdot a_{2}^{8} = 0.\]

\[\left\{ \begin{matrix} a_{1}^{2} - ba_{1} + c = 0 \\ a_{2}^{2} - ba_{2} + c = 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} c = ba_{1} - a_{1}^{2}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ a_{2}^{2} - ba_{2} + ba_{1} - a_{1}^{2} = 0 \\ \end{matrix} \right.\ \]

\[b\left( a_{1} - a_{2} \right) = a_{1}^{2} - a_{2}^{2}\]

\[b = \frac{a_{1}^{2} - a_{2}^{2}}{a_{1} - a_{2}} = a_{1} + a_{2}.\]

\[c = a_{1}\left( a_{1} + a_{2} \right) - a_{1}^{2} = a_{1} \cdot a_{2}.\]