Simplify the polynomial.

The given polynomial is $f(x) = x^3 + 4x + x - 6$.
Combine the like terms:

We look for integer divisors of the constant term $(-6)$, which are $\pm 1, \pm 2, \pm 3, \pm 6$.
Let's test $x=1$:

Since $f(1) = 0$, $x=1$ is a zero of the polynomial, and $(x-1)$ is a factor.

Using synthetic division with the root 1:

1 | 1   0   5   -6
  |     1   1    6
  ----------------
    1   1   6    0

The quotient is $x^2 + x + 6$.
So, $f(x) = (x-1)(x^2 + x + 6)$.

Set the quadratic factor to zero: $x^2 + x + 6 = 0$.
Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=1$, $b=1$, $c=6$.
The discriminant is $\Delta = b^2 - 4ac = (1)^2 - 4(1)(6) = 1 - 24 = -23$.
Since the discriminant is negative, there are no real roots for this quadratic factor. The roots are complex:

The zeros of the polynomial are the real root found in Step 2 and the complex roots found in Step 4.
The zeros are $1, \frac{-1 + i\sqrt{23}}{2}, \frac{-1 - i\sqrt{23}}{2}$.