Question

x^3-2x^2+3x-2=0
x^3+3x^2-3x+4=0
Решите

Answer 1

1)
$x^3-2x^2+3x-2=0 \P(1)=1-2+3-2=0 = textgreater x_1=1 \(x-1)(x^2+ax+b)=x^3+ax^2+bx-x^2-ax-b=\=x^3+x^2(a-1)+x(b-a)-b \ left { {{a-1=-2} atop {b-a=3}} right. \a=1-2=-1 \b=3+a=3-1=2 \(x-1)(x^2-x+2)=0 \x^2-x+2=0 \D textless 0$
Ответ: x=1
2)
$x^3+3x^2-3x+4=0 \P(1)=1+3-3+4=5 \P(-4)=-64+3*16+12+4=48-48=0 = textgreater x_1=-4 \(x+4)(x^2+ax+b)=x^3+ax^2+bx+4x^2+4ax+4b=\=x^3+x^2(a+4)+x(b+4a)+4b \ left { {{a+4=3} atop {4b=4}} right. = textgreater left { {{a=-1} atop {b=1}} right. \(x+4)(x^2-x+1)=0 \x^2-x+1=0 \D textless 0$
Ответ: x=-4

Answer 2

спасибо