Question

分解因式(x四次方+x²-4)(x四次方+x²+3)+10

Answer 1

令a=x^4+x²
原式=(a-4)(a+3)+10
=a²-a-12+10
=a²-a-2
=(a-2)(a+1)
=(x^4+x²-2)(x^4+x²+1)
=(x²+2)(x²烂段扮-1)(x^4+2x²+1-x²)
=(x²+2)(x+1)(x-1)[(x²+1)²-x²]
=(x²+2)(x+1)(x-1)(x²饥灶+x+1)(x²燃庆-x+1)

Answer 2

(x^4+x²-4)(x^4+x²+3)+10
=(x^4+x²)²-(x^4+x²)-12+10
=(x^4+x²培或)²-(x^4+x²)-2
=(x^4+x²-2)(x^4+x²+1)
=(x²+2)(x²-1)(x^4+x²配帆伍轿御+1)
=(x+1)(x-1)(x²+2)(x^4+x²+1)

Answer 3

二楼的答案是正确的。