因式分解 (a-1)(a+1)(a+3)(a+5)+16

分类: 作业答案 • 2022-04-18 13:30:58

问题描述：

答

(a-1)(a+1)(a+3)(a+5)+16
=[(a-1)(a+5)][(a+1)(a+3)]+16
=(a^2+4a-5)(a^2+4a+3)+16
设a^2+4a-5为b
=b(b+8)+16
=b^2+8b+16
十字相乘
=(b+4)^2
=(a^2+4a-1)^2

答

(a-1)(a+1)(a+3)(a+5)+16
=[(a-1)(a+5)][(a+1)(a+3)]+16
=(a^2+4a-5)(a^2+4a+3)+16
=(a^2+4a)^2-2(a^2+4a)-15+16
=(a^2+4a-1)^2
=(a-√5+2)^2(a+√5+2)^2