(x^2+3x+2)(x^2+7x+12)-120
=(x+1)(x+2)(x+3)(x+4)-120
=(x^2+5x+4)(x^2+5x+6)-120
=[(x^2+5x+5)-][(x^2+5x+5)+1]-120
=(x^2+5x+5)^2-1-120
=(x^2+5x+5-11)(x^2+5x+5+11)
=(x^2+5x-6)(x^2+5x+16)
=(x-1)(x+6)(x^2+5x+16)
原式=(x+1)(x=4)*(x+2)(x+3)+1
=(x^2+5x+4)(x^2+5x+6)+1
=(x^2+5x+4)[(x^2+5x+4)+2]+1
=(x^+5x+4)^2+2(x^2+5x+4)+1
=[(x^2+5x+4)+1]^2
=(x^2+5x+5)^2
因为x^2+5x+5=(x+5/2)^2+（5/2根号5)^2>0恒成立,等价于判别式小于零,x^2+5x+5在实数范围内不可再分解,故这就是最后结果.