因式分解的三个问题

(x-1)(x-2)(x-3)(x-4)-24
=[(x-1)(x-4)][(x-2)(x-3)]-24
=[(x^2-5x)+4][(x^2-5x)+6]-24
=(x^2-5x)^2+10(x^2-5x)+24-24
=(x^2-5x)^2+10(x^2-5x)
=(x^2-5x)(x^2-5x+10)
=x(x-5)(x^2-5x+10)

原式=(x-1)(x-4)(x-2)(x-3)-24
=(x^2-5x+4)(x^2-5x+6)-24
令x^2-5x+4=t.
则原式=t(t+2)-24
=t^2+2t-24
=(t-4)(t+6)
代入t得原式=x(x-5)(x^2-5x+10).

=[(X-1)(X-4)][(X-2)(X-3)]-24
=(x^-5x+4)(x^2-5x+6)-24
=(x^2-5x)^2+10(x^2-5x)+24-24
=x(x-5)(x^2-5x+10)

应用整体思想