Đáp án:(Mình đổi đề: $12(x^2-3x-1)=12(x^2+3x-1)$)
Giải thích các bước giải:
$1)(x^2+3x-1)^2-12(x^2+3x-1)+27$$
$=(x^2+3x-1)^2-3(x^2+3x-1)-9(x^2+3x-1)+27$
$=(x^2+3x-1)(x^2+3x-1-3)-9(x^2+3x-1-3)$
$=(x^2+3x-1-9)(x^2+3x-1-3)$
$=(x^2+3x-10)(x^2+3x-4)$
$=(x^2+5x-2x-10)(x^2+4x-x-4)$
$=[x(x+5)-2(x+5)][x(x+4)-(x+4)]$
$=(x+5)(x-2)(x-1)(x+4)$
$2)x^5+x^4-x^3+x^2-x+2$
$=x^5+2x^4-x^4-2x^3+x^3+2x^2-x^2-2x+x+2$
$=x^4(x+2)-x^3(x+2)+x^2(x+2)-x(x+2)+(x+2)$
$=(x+2)(x^4-x^3+x^2-x+1)$
