Question

初二上册代数题,请用换元法来做 因式分解 题目如下

Answer 1

(x+1)(x+2)(x+3)(x+4)-24

=[(x+1)(x+4)][(x+2)(x+3)]-24

=﹙x²＋5x＋4)(x²＋5x＋6﹚-24

设x²+5x=a,换元得

原式

=(a+4)(a+6)-24

=a²+4a+6a+24-24

=a²+10a

=a(a+10)

=(x²＋5x)(x²＋5x＋10)

=x(x+5)(x²＋5x＋10)