Step 1 :The given polynomial is \(P(x)=x^{3}-x^{2}-4 x-6\) and it is given that \(-1-i\) is a zero of the polynomial.
Step 2 :According to the conjugate zeros theorem, if a polynomial has real coefficients and a complex number is a zero of the polynomial, then its conjugate is also a zero of the polynomial. Therefore, the conjugate of \(-1-i\), which is \(-1+i\), is also a zero of the polynomial.
Step 3 :To find the remaining zeros, we can perform polynomial division to divide \(P(x)\) by the quadratic factor corresponding to the pair of complex conjugate zeros. The quotient will be a polynomial of lower degree, and we can find its zeros by setting it equal to zero and solving for \(x\).
Step 4 :After performing the polynomial division, we find that the remaining zero is \(3\).
Step 5 :\(\boxed{\text{The zeros of the polynomial } P(x)=x^{3}-x^{2}-4 x-6 \text{ are } -1-i, -1+i, \text{ and } 3}\)