求解数学题（因式分解）

1)
(x+1)(x+2)(x+3)(x+4)-120
=[(x+1)(x+4)][(x+2)(x+3)]-120
=(x^2+5x+4)(x^2+5x+6)-120
=(x^2+5x)^2+10(x^2+5x)+24-120
=(x^2+5x)^2+10(x^2+5x)-96
=[(x^2+5x)+16][(x^2+5x)-6]
=(x^2+5x+16)(x^2+5x-6)
=(x^2+5x+16)(x+6)(x-1)

2)
(x^2+3x-3)(x^2+3x+4）-8
=[(x^2+3x)-3][(x^2+3x)+4]-8
=(x^2+3x)^2+(x^2+3x)-12-8
=(x^2+3x)^2+(x^2+3x)-20
=(x^2+3x+5)(x^2+3x-4)
=(x^2+3x+5)(x+4)(x-1)

（x+1)(x+2)(x+3)(x+4)-120
=[(x+1)(x+4)][(x+2)(x+3)]-120
=(x^2+5x+4)(x^2+5x+6)-120
令x^2+5x=t
原式=(t+4)(t+6)-120
=t^2+10t+24-120
=t^2+10t-96
=(t+16)(t-6)
=(x^2+5x+16)(x^+5x-6)
=(x^2+5x+16)(x+6)(x-1)

(x^2+3x-3)(x^2+3x+4)-8
令x^2+3x=t
原式=(t-3)(t+4)-8
=t^2+t-20

=(t+5)(t-4)
=(x^2+3x+5)(x^2+3x-4)
=(x^2+3x+5)(x+4)(x-1)
分数给我吧？嘿嘿！

1、（x+1)(x+2)(x+3)(x+4)-120
＝(x+1)(x+4)(x+2)(x+3)-120
＝(x^2+5x