Odpowiedź:
x³-3x²-4x+12=0
x²(x-3) -4(x-3) =0
(x+2)(x-2)(x-3)= 0
x1=-2 x2=2 x3=3
Szczegółowe wyjaśnienie:
[tex]x^{3}-3x^{2}-4x+12 = 0\\\\x^{2}(x-3) - 4(x-3) = 0\\\\(x-3)(x^{2}-4) = 0\\\\(x-3)(x+2)(x-2) = 0\\\\x-3 = 0 \ \vee \ x+2 = 0 \ \vee \ x-2 = 0\\\\x = 3 \ \vee \ x = -2 \ \vee \ x = 2\\\\x \in \{-2, 2, 3\}[/tex]
