3(4+x)(2х-х²) < 0
3(8x-4x²+2x²-x³) < 0
24x-12x²+6x²-3x³ < 0 | :(-3)
-8x+4x²-2x²+x³ < 0
x³+2x²-8x < 0
x(x²+2x-8) < 0
Найдем нули функции:
y = 0
y = x(x²+2x-8)
⇒ x = 0 или x²+2x-8 = 0
D = 4+32 = 36
x₁ =
= -4
x₂ =
=
= 1,5
_______ ______ ______ ___________
-- / + / -- / +
----------- °----------- °------------°-------------------->
//////////// -4 0 /////////// 1,5 х
Ответ: х ∈ (-∞ ; -4) U (0 ; 1,5)