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)
