Use the Factor Theorem to confirm (x-1) is a factor.

According to the Factor Theorem, if $(x-1)$ is a factor of $f(x)$, then $f(1)$ must be equal to 0.
Substitute $x=1$ into the polynomial $f(x) = x^3 + 3x^2 - x - 3$:

Since $f(1)=0$, $(x-1)$ is indeed a factor of $f(x)$.

We can use synthetic division to divide $x^3 + 3x^2 - x - 3$ by $(x-1)$.
The root is $x=1$.

1 | 1   3   -1   -3
  |     1    4    3
  -----------------
    1   4    3    0

The coefficients of the quotient are $1, 4, 3$, and the remainder is 0.
This means that $x^3 + 3x^2 - x - 3 = (x-1)(x^2 + 4x + 3)$.

The quadratic expression is $x^2 + 4x + 3$. We need to find two numbers that multiply to 3 and add to 4. These numbers are 1 and 3.
So, the quadratic factors as $(x+1)(x+3)$.

Now we have the fully factored form of $f(x)$:

To find the zeros, set $f(x)=0$:

This gives us three possible solutions:

The zeros of $f(x)$ are $1, -1,$ and $-3$.