Since the last term of the expression is 16=2^416=24 it's a good idea to try these possible roots: +-1, +-2, +-4, +-8, +-16±1,±2,±4,±8,±16 and its inverses, +-1/2, +-1/4, +-1/8 and +-1/16±12,±14,±18and±116.
Luckily x_1=2x1=2 is a root of the equation:
f(x=2)=2^4-2*2^3-8*2+16=16-16-16+16=0f(x=2)=24−2⋅23−8⋅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+16y=x4−2x3−8x+16
y= x^3*(x-2)-8*(x+2)y=x3⋅(x−2)−8⋅(x+2) => y=(x-2)(x^2-8)y=(x−2)(x2−8)
It remains to factor the polynomial (x^3-8)(x3−8) that has one real root and two complex conjugate roots:
x^3-8=0x3−8=0 => x=(8)^(1/3)x=(8)13
x=2*(cos (k*360^@/3)+i*sin (k*360^@/3))x=2⋅(cos(k⋅360∘3)+i⋅sin(k⋅360∘3)), k=0, 1, 2k=0,1,2
For k=0 -> x_2=2*(cos 0^@+i*sin ^@)=2*(1+i*0)k=0→x2=2⋅(cos0∘+i⋅sin∘)=2⋅(1+i⋅0) => x_2=2x2=2
For k=1 -> x_3=2*(cos 120^@+i*sin 120^@)=2*(-1/2+i*sqrt(3)/2)k=1→x3=2⋅(cos120∘+i⋅sin120∘)=2⋅(−12+i⋅√32) => x_3=-1+sqrt(3)x3=−1+√3
For k=2 -> x_4=2*(cos 240^@+i*sin 240^@)=2*(-1/2-i*sqrt(3)/2)k=2→x4=2⋅(cos240∘+i⋅sin240∘)=2⋅(−12−i⋅√32) => x_3=-1-sqrt(3)x3=−1−√3
So
y=(x-x_1)(x-x_2)(x-x_3)(x-x_4)y=(x−x1)(x−x2)(x−x3)(x−x4)
y=(x-2)(x-2)(x+1-i*sqrt(3))(x+1+i*sqrt(3))y=(x−2)(x−2)(x+1−i⋅√3)(x+1+i⋅√3) => y=(x-2)^2(x+1-i*sqrt (3))(x+1+i*sqrt(3))y=(x−2)2(x+1−i⋅√3)(x+1+i⋅√3)
Or, since (x+1-i*sqrt(3))(x+1+i*sqrt(3))=(x^2+2x+1-(-3))=x^2+2x+4(x+1−i⋅√3)(x+1+i⋅√3)=(x2+2x+1−(−3))=x2+2x+4:
y=(x-2)^2(x^2+2x+4)y=(x−2)2(x2+2x+4)
