You can put this solution on YOUR website! Well from
f(x) = x^3 - 4x^2 + 2x + 1
you can, by inspection, see that x = 1 is one of the zeroes, since f(1) = 0
Now if you factor f(x) you get
f(x) = (x - 1)(x^2 - 3x - 1)
The second polynomial is not factorable but the zeroes can be found using the quadratic formula, so that
x = 1 and
x = (3 ± sqrt(13)) / 2
Thus there are 3 zeroes,
one rational and two irrational...