急!!!!求解两道因式分解题 25分

来源：百度知道 编辑：UC知道 时间：2024/05/30 03:59:57
因式定理法:
(1) X的4次方+2X的3次方-9X的2次方-2X+8

(2) X的4次方+2X的3次方+5X的2次方+4X-12

1.X的4次方+2X的3次方-9X的2次方-2X+8
=(x+1)(x-1)(x+4)(x-2)
X的4次方+2X的3次方-9X的2次方-2X+8
=(X的4次方-X的2次方)+(2X的3次方-2X)-8X的2次方+8

2.X的4次方+2X的3次方+5X的2次方+4X-12
=(x的三次方+5x-6)(x+2)
X的4次方+2X的3次方+5X的2次方+4X-12
=X的4次方+2X的3次方+5X的2次方+10X-6X-12
=X的三次方（X+2）+5X（X+2）-6（X+2）
=x的三次方+5x-6)(x+2)

x^4+2x^3-9x^2-2x+8
==x^4+2x^3-x^2-2x+8-8x^2
=x^2(x^2-1)+2x(x^2-1)-8(x^2-1)
=(x^-1)(x^2+2x-8)
=(x+1)(x-1)(x+4)(x-2)

X^4+2X^3+5X^2+4X-12
=(x^2+5x+2)(x^2+5x+3)-12
=(x^2+5x)^2+5(x^2+5x)+6-12
=(x^2+5x+6)(x^2+5x-1)
=(x+2)(x+3)(x^2+5x-1)

x^4+2x^3-9x^2-2x+8
=x^4+2x^3-3x^2-6x^2-2x+8
=x^2(x^2+2x-3)-2(3x^2+x-4)
=x^2(x+3)(x-1)-2(3x+4)(x-1)
=(x-1)[x^2(x+3)-2(3x+4)]
=(x-1)(x+4)(x-2)(x+1)
2 (x-1)(x-1)(x-2)(x+6)