分解因式:(x-1)(x-2)(x-3)(x-4)-48

如题所述

第1个回答  2006-05-30
解:
(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+6)-24
=(x^2-5x-2)(x^2-5x+12)
(^表示平方)
第2个回答  2006-05-30
(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+6)-24
=(x^2-5x-2)(x^2-5x+12)本回答被提问者采纳
第3个回答  2006-05-30
(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+6)-24
=(x^2-5x-2)(x^2-5x+12)
(^表示平方)
第4个回答  2006-05-30
(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+4)^2+2(x^2-5x+4)-48
=(x^2-5x+4-6)(x^2-5x+4+8)
==(x^2-5x-2)(x^2-5x+12)
第5个回答  2006-06-04
(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+4)^2+2(x^2-5x+4)-48
=(x^2-5x+4-6)(x^2-5x+4+8)
==(x^2-5x-2)(x^2-5x+12)