Find all real and imaginary zeros

y = 11x^3 - 68x^2 +210x -36

It says "some possible zeros are +/- 1, +/- 36, and +/- 2/11"

How are those zeros found? I don't get where they come from.

The possible zeros come from the rational zeros theorem. It says that if a polynomial as rational zeros (it doesn't have to), then for some integers p and q it will have the form p/q where |p| is a factor of the constant term and |q| is a factor of the leading coefficient (absolute value because the root may be negative with neither the constant term nor the leading coefficient being negative). In this case the constant term is -36 which has factors (if we exclude the negative) 1, 2, 3, 4, 6, 9, 12, 18, 36, and the leading coefficient is 11 which has factors 1, 11. So the set of all possible rational zeros is formed by taking every combination of these as positive and negative ratios. Some of these include ±1, ±36/11, ±2/11, ±6/11, etc.

In this case, at most 3 of these could be zero, but none of them could be as well since a polynomial doesn't necessarily have to have rational roots.

Have you studied the rational zeros theorem? It says that if a root of a poynomial is rational and having form p/q then p must be a divisor of the numeric term, in this case -36, and q must be a divisor of the coefficient of the highest order term, in this case 11.

How are those zeros found?
- by the combinations factor of 11 and 36
f(2/11) = ... = 0 ----> (2x-11) is a linear factor , and long division ---> quotient = x^2-6x+10

now solve (2x-11)(x^2-6x+10)=0