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. \( P \) ?

Alright, I need to find all the real zeros of the given polynomial \( P(x) = x^3 - 12x^2 + 16x + 28 \). The problem states that all the real zeros are integers, so I can use the Rational Root Theorem to identify possible rational roots. The Rational Root Theorem suggests that any possible rational root, expressed in lowest terms \( \frac{p}{q} \), has \( p \) as a factor of the constant term and \( q \) as a factor of the leading coefficient. In this case, the constant term is 28, and the leading coefficient is 1. Therefore, the possible rational roots are the factors of 28, which are \( \pm1, \pm2, \pm4, \pm7, \pm14, \pm28 \). I'll start by testing these possible roots by substituting them into the polynomial to see if they yield zero. 1. **Testing \( x = 1 \):** \[ P(1) = 1^3 - 12(1)^2 + 16(1) + 28 = 1 - 12 + 16 + 28 = 33 \neq 0 \] So, \( x = 1 \) is not a root. 2. **Testing \( x = -1 \):** \[ P(-1) = (-1)^3 - 12(-1)^2 + 16(-1) + 28 = -1 - 12 - 16 + 28 = -1 \neq 0 \] So, \( x = -1 \) is not a root. 3. **Testing \( x = 2 \):** \[ P(2) = 2^3 - 12(2)^2 + 16(2) + 28 = 8 - 48 + 32 + 28 = 20 \neq 0 \] So, \( x = 2 \) is not a root. 4. **Testing \( x = -2 \):** \[ P(-2) = (-2)^3 - 12(-2)^2 + 16(-2) + 28 = -8 - 48 - 32 + 28 = -60 \neq 0 \] So, \( x = -2 \) is not a root. 5. **Testing \( x = 4 \):** \[ P(4) = 4^3 - 12(4)^2 + 16(4) + 28 = 64 - 192 + 64 + 28 = -36 \neq 0 \] So, \( x = 4 \) is not a root. 6. **Testing \( x = -4 \):** \[ P(-4) = (-4)^3 - 12(-4)^2 + 16(-4) + 28 = -64 - 192 - 64 + 28 = -292 \neq 0 \] So, \( x = -4 \) is not a root. 7. **Testing \( x = 7 \):** \[ P(7) = 7^3 - 12(7)^2 + 16(7) + 28 = 343 - 588 + 112 + 28 = -105 \neq 0 \] So, \( x = 7 \) is not a root. 8. **Testing \( x = -7 \):** \[ P(-7) = (-7)^3 - 12(-7)^2 + 16(-7) + 28 = -343 - 588 - 112 + 28 = -1015 \neq 0 \] So, \( x = -7 \) is not a root. 9. **Testing \( x = 14 \):** \[ P(14) = 14^3 - 12(14)^2 + 16(14) + 28 = 2744 - 2352 + 224 + 28 = 644 \neq 0 \] So, \( x = 14 \) is not a root. 10. **Testing \( x = -14 \):** \[ P(-14) = (-14)^3 - 12(-14)^2 + 16(-14) + 28 = -2744 - 2352 - 224 + 28 = -5302 \neq 0 \] So, \( x = -14 \) is not a root. 11. **Testing \( x = 28 \):** \[ P(28) = 28^3 - 12(28)^2 + 16(28) + 28 = 21952 - 9408 + 448 + 28 = 13016 \neq 0 \] So, \( x = 28 \) is not a root. 12. **Testing \( x = -28 \):** \[ P(-28) = (-28)^3 - 12(-28)^2 + 16(-28) + 28 = -21952 - 9408 - 448 + 28 = -31772 \neq 0 \] So, \( x = -28 \) is not a root. Since none of the possible rational roots satisfy the equation \( P(x) = 0 \), it implies that the polynomial does not have any real integer zeros. Therefore, the polynomial \( P(x) = x^3 - 12x^2 + 16x + 28 \) does not have any real zeros that are integers.