
分解因式:(x4+x2-4)(x4+x2+3)+10=______
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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)
=(x2+x+1)(x2-x+1)((x2+2)(x+1)(x-1)
故答案为:(x2+x+1)(x2-x+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)
=(x2+x+1)(x2-x+1)((x2+2)(x+1)(x-1)
故答案为:(x2+x+1)(x2-x+1)((x2+2)(x+1)(x-1).
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