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.) \[ P(x)=x^{3}-13 x^{2}-16 x+28 \] You do not have the correct number of terms in your comma-separated list. Write thelpolynomial in factored form.

We start with the polynomial   P(x) = x³ − 13x² − 16x + 28. Step 1. Find a zero using the Rational Root Theorem. The possible rational zeros are the factors of 28: ±1, ±2, ±4, ±7, ±14, ±28. Test x = 1:   P(1) = 1 − 13 − 16 + 28 = 0. Since P(1) = 0, x = 1 is a zero. Step 2. Factor (x − 1) from P(x) by dividing: Dividing P(x) by (x − 1) (using synthetic or long division) gives the quadratic quotient:   x² − 12x − 28. Step 3. Factor the quadratic. Solve x² − 12x − 28 = 0 using the quadratic formula:   x = [12 ± √(144 + 112)] / 2 = [12 ± √256] / 2 = [12 ± 16] / 2. Thus:   x = (12 + 16)/2 = 14  and  x = (12 − 16)/2 = –2. So the zeros are x = 1, x = 14, and x = –2 (each with multiplicity 1). Step 4. Write the polynomial in factored form. The factored form is:   P(x) = (x − 1)(x − 14)(x + 2). Answer in comma-separated list (using appropriate multiplicities):   1, 14, -2