The polynomial given in the image is:

$P(x) = x^4 - 3x^3 - 27x^2 - 3x - 28$

To find all zeros (roots) of this polynomial, we can follow these steps:

1. Use the Rational Root Theorem to test for possible rational roots. According to this theorem, the potential rational roots are the factors of the constant term (here, -28) divided by the factors of the leading coefficient (here, 1). This gives possible roots: ±1, ±2, ±4, ±7, ±14, ±28.

2. Test these potential roots using synthetic division or direct substitution to see if any of them are actual roots of the polynomial.

3. Once a root is found, factor it out and reduce the polynomial's degree, then solve for the remaining roots by factoring or using the quadratic formula.

Let me begin by testing the possible rational roots and solving this equation.The polynomial $P(x) = x^4 - 3x^3 - 27x^2 - 3x - 28$ has two rational roots: $x = -4$ and $x = 7$.

Next, I will factor these roots out of the polynomial and solve the remaining quadratic equation to find the other zeros. Let's proceed with that.After dividing the polynomial by $(x + 4)$ and $(x - 7)$, the remaining quadratic polynomial is:

$x^2 + 1$

The solutions to this quadratic equation are the complex roots of the polynomial:

$x^2 + 1 = 0 \quad \Rightarrow \quad x = \pm i$

Thus, the complete list of zeros (including repetitions) of the polynomial $P(x) = x^4 - 3x^3 - 27x^2 - 3x - 28$ is:

$x = -4, \, 7, \, i, \, -i$

Would you like more details on any of these steps? Here are some related questions:

1. How do you use synthetic division to factor polynomials?
2. What is the Rational Root Theorem and how does it work?
3. How do complex roots arise from quadratic equations?
4. Can all polynomials be factored into real numbers?
5. What are some techniques for solving higher-degree polynomials?

Tip: Always check for both real and complex roots when solving higher-degree polynomials!