题目内容
分解因式:(x4+x2-4)(x4+x2+3)+10=
(x4+x2+1)(x2+2)(x+1)(x-1)
(x4+x2+1)(x2+2)(x+1)(x-1)
.分析:首先利用换元,令x4+x2=y,然后根据十字相乘法进行因式分解,最后再将x4+x2=y,代入进行还原,得出结果.
解答:解:令x4+x2=y,
∴原式=(y-4)(y+3)+10
=y2-y-2
=(y+1)(y-2)
将x4+x2=y代入,
所以原式=(x4+x2+1)(x4+x2-2)
=(x4+x2+1)(x2+2)(x2-1)
=(x4+x2+1)(x2+2)(x+1)(x-1).
故答案为为(x4+x2+1)(x2+2)(x+1)(x-1).
∴原式=(y-4)(y+3)+10
=y2-y-2
=(y+1)(y-2)
将x4+x2=y代入,
所以原式=(x4+x2+1)(x4+x2-2)
=(x4+x2+1)(x2+2)(x2-1)
=(x4+x2+1)(x2+2)(x+1)(x-1).
故答案为为(x4+x2+1)(x2+2)(x+1)(x-1).
点评:本题综合考查了十字相乘法和换元法,做这类题必须要记得还原回去,不能得出的结果为(y+1)(y-2).
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