Х(х-1)(х+2)(х+3)>0(x^2 - x)(x^2 + 3x + 2x + 6)>0
(x^2 - x)(x^2 + 5x + 6)>0
x^2 - x = 0
x^2 = x
x=1 x=0
x^2 + 5x + 6 = 0
D = 25 - 24 = 1
x = (-5 - 1)\2 = -6\2 = -3
x= (-5 + 1)\2 = -4\2 = -2
+ - + - +
----- • ---- • ----------- • --------- • ----->x
-3 -2 0 1
Ответ: (-∞ ; -3)(-2;0)(1; +∞)