Quick math problem question

In general, how do you locate all the zeros of a polynomial?

Here's a problem from my practice test:
Locate all zeros of this polynomial:
y = 4x^3 -12x^2 - 4x + 12

Thanks.

y = 4x^2(x-3) - 4(x-3)
y = (x-3)(4x^2-4) = 4(x-3)(x+1)(x-1)
Zeros are 1, 3 and -1

y = 4x^3 -12x^2 - 4x + 12
0 = 4(x^3 – 3x^2 – x + 3)
the possibility of making zero to a value of 3 is ±{1,3}. There are 4 values. check one by one and I get the value of x = 1 or (x – 1)
means (x - 1) is one of the main factors. use as a divider
(x^3 – 3x^2 – x + 3)/(x – 1) = x² - 2x – 3
Then
(x – 1)(x² - 2x – 3)
for (x² - 2x – 3) by factoring, then
(x - 3)(x + 1), then
(x – 1)(x – 3)(x + 1)
and obtained zero-makers
x = 1
x = 3
x = -1

use synthetic division

y = 4 (x-3) (x+1) (x-1) --> 1, 3 and -1