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

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