Question

因式分解:(a+1)(a+2)(a+3)(a+4)-120

Answer 1

答：

原式

=(a+1)(a+4)(a+2)(a+3)-120

=(a^2+5a+4)(a^2+5a+6)-120

=(a^2+5a)^2+10(a^2+5a)+24-120

=(a^2+5a)^2+10(a^2+5a)-96

=(a^2+5a+16)(a^2+5a-6)

=(a+6)(a-1)(a^2+5a+16)