Question

Найдите действительные корни уравнения! Срочно!
х^3-3х^2-6х+8=0

Answer 1

(х³+8)-(3х²+6х)=0

(х+2)(х²-2х+4)-3х(х+2)=0

(х+2)(х²-2х+4-3х)=0

(х+2)(х²-5х+4)=0

х+2=0  или      х²-5х+4

х₁=-2              по Т. Виета х₂=1, х₃=4

Answer 2

$x {}^{3} - 3x {}^{2} - 6x + 8 = 0 \ (x + 2)(x {}^{2} - 2x + 4) - 3x(x + 2) = 0 \ (x + 2)(x {}^{2} - 2x + 4 - 3x) = 0 \ (x + 2)(x {}^{2} - 5x + 4) = 0 \ (x + 2)(x {}^{2} - x - 4x + 4) = 0 \ (x + 2)(x(x - 1) - 4(x - 1)) = 0 \ (x + 2)(x - 1)(x - 4) = 0 \ x + 2 = 0 : : : x - 1 = 0 : : x - 4 = 0$
х=-2
х=1
х=4