分解因式(x-1)(x+2)(x-3)(x+4)+24

这需要技巧，要把四个（）分成两组，诀窍是：两组之积中，x的一次项相同。比如这道题的x²+x-2和x²+x-12
原式=(x-1)(x+2)*(x-3)(x+4)+24
=(x²+x-2)(x²+x-12)+24
换元，设x²+x-2=t
=t(t-10)+24
=t²-10t+24
=(t-4)(t-6)
代回
=(x²+x-6)(x²+x-8)
=(x+3)(x-2)(x²+x-8)

(x-1)(x+2)(x-3)(x+4)+24
=(x²+x-2)(x²+x-12)+24
=(x²+x)²-14(x²+x)+24+24
=(x²+x)²-14(x²+x)+48
=(x²+x-6)(x²+x-8)
=(x+3)(x-2)(x²+x-8)
希望能帮到你O(∩_∩)O

(x-1)(x+2)(x-3)(x+4)+24
=(x²+x-2)(x²+x-12)+24
=(x²+x)²-12(x²+x)-2(x²+x)+24+24
=(x²+x)²-14(x²+x)+48
=(x²+x-6)(x²+x-8)
=(x+3)(x-2)(x²+x-8)

(x-1)(x+2)(x-3)(x+4)+24
=(x^2+x-2)(x^2+x-12)+24
=(x^2+x)^2-14(x^2+x)+48
=(x^2+x-6)(x^2+x-8)
=(x-2)(x+3)(x^2+x-8)