初三数学因式分解题

(x²+3x-3)(x²+3x+4)-8
=(x^2+3x)^2＋(x^2+3x)－20
=(x^2+3x-4)(x^2+3x＋5)
=(x-1)(x+4)(x^2+3x＋5)

x(x + 1)(x - 1)(x + 2) - 24
= x(x^2 - 1)(x + 2) - 24
= x(x^3 + 2x^2 - x - 2) - 24
= x^4 + 2x^3 - x^2 - 2x - 24
= (x^4 - 2x^3) + 4x^3 - x^2 - 2x - 24
= x^3(x - 2) + 4x^3 - 8x^2 + 7x^2 - 2x - 24
= (x^3 + 4x^2)(x - 2) + (7x^2 - 14x) + (12x - 24)
= (x^3 + 4x^2 + 7x + 12)(x - 2)
= [(x^3 + 3x^2) + (x^2 + 3x) + (4x + 12)](x - 2)
= (x^2 + x + 4)(x + 3)(x - 2)

2x²+xy-y²-4x+5y-6
=2x^2+(y-4)x+(-y^2+5y-6)
=2x^2+(y-4)x+[-(y^2-5y+6)]
=2x^2+(y-4)x+[-(y-2)(y-3)]
=[1x+(y-3)][2x+(-y+2)]
=(x+y-3)(2x-y+2)

1 (x²+3x-3)(x²+3x+4)-8=
令x²+3x=K，则
上式=(K-3)(K+4)-8
=K²+K-20
=(K+5)(K-4)
所以原式
=(x²+3x+5)(x²+3x-4)
=(x+4)(x-1)(x²+3x+5)

2 (x-1)x(x+1)(x+2)-24
=x(x+1)(x-1)(x+2)-24
=(x²+x)(x²+