用十字相乘法解：(x2+3x-3)(x2+3x+4)-8

分类: 作业答案 • 2022-05-04 23:40:01

问题描述：

答

设x2+3x=t
(x2+3x-3)(x2+3x+4)-8
=(t-3)(t+4)-8
=t^2+t-12-8
=t^2+t-20
=(t+5)(t-4)
=(x2+3x+5)(x2+3x-4)
=(x2+3x+5)(x+4)(x-1)搞错了是这道题2x2+xy-y2-4x+5y-6