Find all other zeros of \( P(x)=x^{3}+6 x^{2}+18 x+40 \), given that \( -1-3 i \) is a zer (If there is more than one zero, separate them with commas.)

To find all the zeros of the polynomial \( P(x) = x^3 + 6x^2 + 18x + 40 \), given that one zero is \( -1 - 3i \), we first use the fact that the coefficients of the polynomial are real. This implies that complex roots must come in conjugate pairs. Thus, the other zero must be \( -1 + 3i \). Next, we can factor \( P(x) \) by finding the polynomial formed by these two zeros. The sum and product of the roots \( -1 - 3i \) and \( -1 + 3i \) can be computed as follows: 1. The sum of the roots: \[ (-1 - 3i) + (-1 + 3i) = -2 \] 2. The product of the roots: \[ (-1 - 3i)(-1 + 3i) = (-1)^2 - (3i)^2 = 1 - (-9) = 10 \] Thus, the quadratic corresponding to these complex roots is \[ x^2 + 2x + 10. \] Now, we divide \( P(x) \) by \( x^2 + 2x + 10 \) to find the remaining zero. Using synthetic or polynomial division, we find: \[ P(x) \div (x^2 + 2x + 10) = x + 4. \] So, \( P(x) \) can be factored as: \[ P(x) = (x + 4)(x^2 + 2x + 10). \] Setting \( x + 4 = 0 \) gives us the remaining zero: \[ x = -4. \] Thus, the full list of zeros for \( P(x) \) is: \[ -1 - 3i, -1 + 3i, -4. \] The complete set of zeros is: \[ -1 - 3i, -1 + 3i, -4. \]