Find all zeros of the polynomial. (Enter your answers as a comma-separated list. Enter all answers including repetitions.) \[ P(x)=x^{3}+5 x^{2}+9 x+45 \] \( x=\square \)

To find the zeros of the polynomial \( P(x) = x^3 + 5x^2 + 9x + 45 \), we can start by applying the Rational Root Theorem, testing potential rational roots such as ±1, ±3, ±5, ±9, ±15, ±45. After testing these roots, we find that \( x = -3 \) is a root. We can then perform synthetic division of \( P(x) \) by \( (x + 3) \) to simplify the polynomial: 1. \( P(x) \div (x + 3) \) gives us \( x^2 + 2x + 15 \). 2. Now we solve \( x^2 + 2x + 15 = 0 \) using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here \( a = 1, b = 2, c = 15 \): - Calculating the discriminant: \( b^2 - 4ac = 2^2 - 4(1)(15) = 4 - 60 = -56 \). Since the discriminant is negative, we will have complex roots. The solutions are: - \( x = \frac{-2 \pm \sqrt{-56}}{2} = -1 \pm i\sqrt{14} \). So the zeros of the polynomial \( P(x) \) are: \[ x = -3, -1 + i\sqrt{14}, -1 - i\sqrt{14}. \] Thus, enter your answers as: \[ x = -3, -1 + i\sqrt{14}, -1 - i\sqrt{14}. \]