ОГЭ по математике. №20 Уравнения. ФИПИ. Задача 58

ОТВЕТ:  \(-4;\,\,\,\,-3;\,\,\,\,4.\)

\({x^3} + 3{x^2} = 16x + 48\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,{x^3} + 3{x^2}-16x-48 = 0.\)

Разложим левую часть полученного уравнения на множители методом группировки:

\({x^3} + 3{x^2}-16x-48 = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,{x^2}\left( {x + 3} \right)-16\left( {x + 3} \right) = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left( {x + 3} \right)\left( {{x^2}-16} \right) = 0\,\,\,\,\,\, \Leftrightarrow \)

\( \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x + 3 = 0,\\{x^2}-16 = 0\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x = -3,\\\left( {x-4} \right)\left( {x + 4} \right) = 0\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x = -3,\\x-4 = 0,\\x + 4 = 0\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x = -3,\\x = 4,\\x = -4.\end{array} \right.\)

Ответ:  \(-4;\,\,\,\,-3;\,\,\,\,4.\)