Since the last term of the expression is #16=2^4# it's a good idea to try these possible roots: #+-1, +-2, +-4, +-8, +-16# and its inverses, #+-1/2, +-1/4, +-1/8 and +-1/16#.
Luckily #x_1=2# is a root of the equation:
#f(x=2)=2^4-2*2^3-8*2+16=16-16-16+16=0#
If at first this case of factorization (common factor) was not immediately seen, now we can do as follows
#y= x^4-2x^3-8x+16#
#y= x^3*(x-2)-8*(x+2)# => #y=(x-2)(x^2-8)#
It remains to factor the polynomial #(x^3-8)# that has one real root and two complex conjugate roots:
#x^3-8=0# => #x=(8)^(1/3)#
#x=2*(cos (k*360^@/3)+i*sin (k*360^@/3))#, #k=0, 1, 2#
For #k=0 -> x_2=2*(cos 0^@+i*sin ^@)=2*(1+i*0)# => #x_2=2#
For #k=1 -> x_3=2*(cos 120^@+i*sin 120^@)=2*(-1/2+i*sqrt(3)/2)# => #x_3=-1+sqrt(3)#
For #k=2 -> x_4=2*(cos 240^@+i*sin 240^@)=2*(-1/2-i*sqrt(3)/2)# => #x_3=-1-sqrt(3)#
So
#y=(x-x_1)(x-x_2)(x-x_3)(x-x_4)#
#y=(x-2)(x-2)(x+1-i*sqrt(3))(x+1+i*sqrt(3))# => #y=(x-2)^2(x+1-i*sqrt (3))(x+1+i*sqrt(3))#
Or, since #(x+1-i*sqrt(3))(x+1+i*sqrt(3))=(x^2+2x+1-(-3))=x^2+2x+4#:
#y=(x-2)^2(x^2+2x+4)#
