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.

To find the real zeros of the polynomial \( P(x) = x^{3} - 13x^{2} - 16x + 28 \), we can apply the Rational Root Theorem. Testing possible rational roots, it turns out \( x = 1 \) is a root. Dividing the polynomial \( P(x) \) by \( (x - 1) \) leads to \( P(x) = (x - 1)(x^2 - 12x - 28) \). Next, we need to find the roots of \( x^2 - 12x - 28 = 0 \) using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Plugging in \( a = 1 \), \( b = -12 \), and \( c = -28 \): \[ x = \frac{12 \pm \sqrt{(-12)^2 - 4(1)(-28)}}{2(1)} = \frac{12 \pm \sqrt{144 + 112}}{2} = \frac{12 \pm \sqrt{256}}{2} = \frac{12 \pm 16}{2} \] This gives us the two solutions: \[ x = \frac{28}{2} = 14 \quad \text{and} \quad x = \frac{-4}{2} = -2 \] Thus, the real zeros are: \( 1, 14, -2 \). The factored form of the polynomial is: \[ P(x) = (x - 1)(x - 14)(x + 2) \]