ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 309

\[\boxed{\mathbf{309}.}\]

\[1)\ P(x) = x^{3} + 5x^{2} + 11x + 7;\ \]

\[\ a = - 1:\]

\[x^{2} + 4x + 7 = 0\]

\[D_{1} = 4 - 7 = - 3 < 0\]

\[нет\ корней.\]

\[P(x) = (x + 1)(x^{2} + 4x + 7.\]

\[2)\ P(x) = 3x^{3} + 10x^{2} + 4x + 3;\ \ \]

\[a = - 3:\]

\[3x^{2} + x + 1 = 0\]

\[D = 1 - 12 = - 11 < 0\]

\[нет\ корней.\]

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

\[3)\ P(x) = x^{3} + 4x^{2} - 7x - 10;\ \ \]

\[a = - 5:\]

\[x^{2} - x - 2 = 0\]

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

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

\[P(x) = (x + 5)(x + 1)(x - 2).\]

\[4)\ P(x) = x^{3} - 2x^{2} - 5x + 10;\]

\[\ \ a = 2:\]

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

\[x^{2} = 5\]

\[x = \pm \sqrt{5}.\]

\[P(x) =\]

\[= \left( x + \sqrt{5} \right)(x - 2)\left( x - \sqrt{5} \right).\]