已知3x^2+12x+1=0,那么多项式(x-1)(x+1)(x+3)(x+5)的值为

3x^2+12x+1=0
∴3x²+12x=-1
x²+4x=-1/3
（x-1）（x+1）（x+3）(x+5)
=(x-1)(x+5)(x+1)(x+3)
=(x²+4x-5)(x²+4x+3)
=(-1/3 -5)×(-1/3 +3)
=-16/3×8/3
=-128/9