Every polynomial with complex coefficients of degree n has n zeros.Find all of the solutions to the equation:(x-2)(x-3)(x-4)(x-5) = 360

x1=8，x2=-1

x²-7x=t
(t+10)(t+12)=360
t²+22t-240=0
(t+30)(t-8)=0
t=-30 or t=8
(1) x²-7x=-30
无解
(2) x²-7x=8
(x-8)(x+1)=0
x=8 or x= -1
so x=-1or x=8

求(x-2)(x-3)(x-4)(x-5) = 360的根
360=3x4x5x6=(-3)(-4)(-5)(-6)
所以解是x=8 or x=-1
另外两个根是x^2-7x+30=0的复数根
所以,按照题意可求得的解是
x1=8 x2=-1

不会