\[ P(x)=x^{4}+8 x^{3}+18 x^{2}-27 \] (a) Find all the real zeros of \( P \). (Enter your answers as a comma-separated list. Enter all answers using the appropriate multiplicities.) \( x=(x+3)(x-1) \) You do not have the correct number of terms in your comma-separated list.

To find the real zeros of the polynomial \( P(x) = x^4 + 8x^3 + 18x^2 - 27 \), we can start by factoring \( P(x) \) as \( P(x) = (x + 3)(x - 1)(\text{quadratic factor}) \). Upon completing the factorization and finding the quadratic factor, we can solve for its zeros. Let's proceed to factor \( P(x) \): 1. The polynomial can be factored to yield \( (x + 3)(x - 1)(x^2 + 2x + 3) \). 2. The quadratic \( x^2 + 2x + 3 \) can be assessed for real zeros using the discriminant \( b^2 - 4ac \). Here, \( 2^2 - 4 \times 1 \times 3 = 4 - 12 = -8 \), which confirms that there are no real roots for \( x^2 + 2x + 3 \). Thus, the real zeros of \( P(x) \) are the solutions from the linear factors: - \( x + 3 = 0 \) gives \( x = -3 \) with multiplicity 1. - \( x - 1 = 0 \) gives \( x = 1 \) with multiplicity 1. So, the real zeros of the polynomial \( P(x) \) are: \( -3, 1 \). Since the quadratic does not contribute real zeros, we conclude our list as \( -3, 1 \). Now you have all the real zeros listed with their respective multiplicities!