算道题目

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

解:
原式=[(x+1)(x+4)]*[(x+2)(x+3)]-24
=(x^2+5x+4)*(x^2+5x+6)-24
设A=x^2+5x,则
原式=(A+4)(A+6)-24=A^2+10A+24-24=A^2+10A
原式=A(A+10)
=(x^2+5x)(x^2+5x+10)
=x(x+5)(x^2+5x+10)

(x^2-7x+6)(x^2-x-6)+56
=(x-6)(x-1)(x-3)(x+2)+56
=[(x-6)(x+2)][(x-1)(x-3)]+56
=(x^2-4x-12)(x^2-4x+3)+56
=(x^2-4x-12)^2+15(x^2-4x-12)+56
=(x^2-4x-12+7)(x^2-4x-12+8)
<十字相乘法>
=(x^2-4x-5)(x^2-4x-4)
=(x-5)(x+1)(x^2-4x-4)