因式分解（x+1)(x+2)(x+3)(x+4)-48

Question

如题（x+1)(x+2)(x+3)(x+4)-48

需详细解题思路和步骤，最好能列明是用的什么方法，谢谢

Answer 1

（x+1)(x+2)(x+3)(x+4)-48
=[（x+1)(x+4)]*[(x+2)(x+3)]-48
=(x^2+5x+4)(x^2+5x+6)-48
=(x^2+5x)^2+10(x^2+5x)-24
=(x^2+5x+12)(x^2+5x-2)

解此题时x^2+5x是关键.本回答被提问者采纳

Answer 2

[(x+1)(x+4)][(x+2)(x+3)]-48
=(x^2+5x+4)(x^2+5x+6)-48
=(x^2+5x)^2+10(x^2+5x)-24
=(x^2+5x+12)(x^2+5x-2)

Answer 3

=（x+1）（x+4）（x+2）（x+3）-48
=(x^2+5x+4)(x^2+5x+6)-48
=(x^2+5x)^2+10(x^2+5x)+24-48
=(x^2+5x)^2+10(x^2+5x)-24
=(x^2+5x-2)(x^2+5x+12)