Решить уравнение

(х-2) (х²+2 х+1) = 4 (х+1)

(х-2) (х²+2 х+1) = 4 (х+1)

(х-2) (х+1) ² - 4 (х+1) = 0

(х+1) ((х-2) (х+1) - 4) = 0

х+1 = 0 (х-2) (х+1) - 4 = 0

х1 = - 1 х² - х - 6 = 0

по т. Виетта

х2 = - 2, х3 = 3

(х-2) (х²+2 х+1) = 4 (х+1)

(х-2) (х+1) ²-4 (х+1) = 0

(х+1) ((х-2) (х+1) - 4) = 0

(х+1) (х²-2 х-2+х-4) = 0

(х+1) (х²-х-6) = 0

х+1=0

х=-1

или

х²-х-6=0

D=1+24=25

x = (1-5) : 2=-2

x = (1+5) : 2=3

ответ: - 2; -1; 3