Using Zero Rational Theorem you see that the potential roots p/qpq are {1,2,4,8,16}{1,2,4,8,16}.
First root that we will try will x-4x−4
(x^3 - 6x^2 + 4x + 16)/(x-4)x3−6x2+4x+16x−4 apply long division
4 |bar(1 - 6 + 4 + 164∣¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯1−6+4+16
" " |1 + 4 - 8 - 16 ∣1+4−8−16
" " |1 -2 - 4 - 0 ∣1−2−4−0
Thus (x-4)(x−4) divides our polynomial without remainder and :. it is a factor.
(x^3 - 6x^2 + 4x + 16)=color(brown)((x^2-2x-4))(x-4)
Now factor the binomial color(brown)(x^2-2x-4)
Use quadratic formula:
x_(1,2)=-(b+-sqrt(b^2-4ac))/(2a)=(2+- sqrt((-2)^2-4(1)(-4) ))/2
x_(1,2)=(2+- sqrt(4+16))/2=(2+- 2sqrt(5))/2=1+-sqrt(5)
So the roots or zeros are: x_(1,2)=1+-sqrt(5) and x=4
