分解因式(x^2-x)^2-8x^2+8x+12

分类: 作业答案 • 2022-07-14 18:46:42

问题描述：

答

(x&sup2-x)&sup2-8x&sup2+8x+12
=(x&sup2-x)&sup2-8(x&sup2-x)+12
=[(x&sup2-x)-2][(x&sup2-x)-6]
=(x&sup2-x-2)(x&sup2-x-6)
=(x+1)(x-2)(x+2)(x-3)