2 x^2-4 x+1=0,(x+1)(x+2)(x-3)(x-4)=?

2 x^2-4 x+1=0,(x+1)(x+2)(x-3)(x-4)=?

2 x&菗178;-4 x=-1
x&菗178;-2 x=-1/2
原式=[(x+1)(x-3)][(x+2)(x-4)]
=(x&菷178;-2 x-3)(x&菗178;-2 x-8)
=(-1/2-3)(-1/2-8)
=119/4