All the real zeros of the given polynomial are integers. Find the zeros. (Enter your answers as a comma-separated list. Enter all answers using the appropriate multiplicities.) \[ P(x)=x^{3}+4 x^{2}-3 x-18 \] Write the polynomial in factored form. \( P(x)=\square \)

To find the real zeros of the polynomial \( P(x) = x^3 + 4x^2 - 3x - 18 \), we can use the Rational Root Theorem, which suggests that possible rational zeros are the factors of the constant term (-18) divided by the factors of the leading coefficient (1). The possible rational roots are ±1, ±2, ±3, ±6, ±9, ±18. By testing these potential zeros, we find that \( x = 2 \) is a root since \( P(2) = 2^3 + 4(2)^2 - 3(2) - 18 = 8 + 16 - 6 - 18 = 0 \). We can then use synthetic division to factor the polynomial \( P(x) \) by \( (x - 2) \). After performing synthetic division, we get: \[ P(x) = (x - 2)(x^2 + 6x + 9) \] The quadratic \( x^2 + 6x + 9 \) can be factored further: \[ x^2 + 6x + 9 = (x + 3)^2 \] Thus, we can express \( P(x) \) as: \[ P(x) = (x - 2)(x + 3)^2 \] The real zeros of the polynomial are 2 and -3 (with multiplicity 2). Therefore, the final answers are: Zeros: \( 2, -3, -3 \) Factored form: \[ P(x) = (x - 2)(x + 3)^2 \]