Question

Найдите корни многочлена x^3-3x^2-6x+8

Answer 1

Ответ:

x1=-2, x2=1, x3=4

Объяснение:

$x^3-3x^2-6x+8=0\\(x+2)*(x^2-2x+4)-3x*(x+2)=0\\(x+2)*(x^2-2x+4-3x)=0\\(x+2)*(x^2-5x+4)=0\\(x+2)*(x^2-x-4x+4)=0\\(x+2)*(x*(x-1)-4*(x-1))=0\\(x+2)*(x-1)*(x-4)=0\\x+2=0\\x=-2\\x-1=0\\x=1\\x-4=0\\x=4$