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把下列各式分解因式:(1)x3+x2y-xy2-y3;(2)(a2+2)(a2+1)-30.

题目详情
把下列各式分解因式:
(1)x3+x2y-xy2-y3
(2)(a2+2)(a2+1)-30.
▼优质解答
答案和解析
(1)x3+x2y-xy2-y3
=(x3+x2y)−(xy2+y3)
=x2(x+y)−y2(x+y)
=(x+y)(x2−y2)
=(x+y)(x+y)(x−y)
=(x+y)2(x−y)


(2)(a2+2)(a2+1)-30
=a4+3a2+2−30
=a4+3a2−28
=(a2−4)(a2+7)
=(a+2)(a−2)(a2+7)