All the real zeres of the given polynemial are integers. Find the zeros. (Enter your answers as a comma-separated list. Enter all answers using the appropriate multiplicities \[ P(x)=x^{3}-12 x^{2}=16 x+28 \] \( P(x)=\square \) Write the polynemial in factored form. \( P \) ?

Let’s start by rearranging the given polynomial equation \( P(x) = x^{3} - 12x^{2} - 16x - 28 \) to the standard form: \[ P(x) = x^{3} - 12x^{2} - 16x - 28 \] To find the rational roots, you can use the Rational Root Theorem. First, we can test some integer values. If we let \( x = 14 \): \[ P(14) = 14^{3} - 12(14)^{2} - 16(14) - 28 = 0 \] Thus, one root is \( x = 14 \). Now we can perform synthetic division of \( P(x) \) by \( x - 14 \): After the division, we find the other factorable polynomial is \( (x - 14)(x^2 + 2) \). Setting \( x^2 + 2 = 0 \) gives us the non-real roots \( x = i\sqrt{2} \) and \( x = -i\sqrt{2} \). So the only integer zero of the polynomial \( P(x) \) is: **Zeros: 14 (with multiplicity 1)** In factored form, the polynomial is: **\( P(x) = (x - 14)(x^2 + 2) \)**