Найдите корни уравнения x(x+1)(x+2)(x+3)=120.

Ответ:

\[x(x + 1)(x + 2)(x + 3) = 120\]

\[\left( x(x + 3) \right)\left( (x + 1)(x + 2) \right) = 120\]

\[\left( x^{2} + 3x \right)\left( x^{2} + 3x + 2 \right) = 120\]

\[y = x^{2} + 3x:\]

\[y(y + 2) = 120\]

\[y^{2} + 2y - 120 = 0\]

\[D = 1 + 120 = 121\]

\[y_{1} = - 1 + 11 = 10;\ \ \]

\[y_{2} = - 1 - 11 = - 12\]

\[1)\ y = - 12:\]

\[x^{2} + 3x = - 12\]

\[x^{2} + 3x + 12 = 0\]

\[D = 9 - 48 < 0\]

\[нет\ корней.\]

\[2)\ y = 10:\]

\[x^{2} + 3x = 10\]

\[x^{2} + 3x - 10 = 0\]

\[x_{1} + x_{2} = - 3;\ \ x_{1} \cdot x_{2} = - 10\]

\[x_{1} = - 5;\ \ x_{2} = 2.\]

\[Ответ:x = - 5;x = 2.\]