The polynomial function in the question is $f(x) = x^3 + 4x^2 + x + 4$. We need to find the zeros of this polynomial, i.e., the values of $x$ that make $f(x) = 0$.

Step-by-step solution:

1. Attempt Rational Root Test: We can apply the Rational Root Theorem, which suggests that any rational root must be a factor of the constant term (which is $4$) divided by a factor of the leading coefficient (which is $1$).

   The possible rational roots are $\pm 1, \pm 2, \pm 4$.

2. Test possible roots: Let's start testing these values by plugging them into the polynomial to see if any of them satisfy $f(x) = 0$.

   • $f(1) = 1^3 + 4(1)^2 + 1 + 4 = 1 + 4 + 1 + 4 = 10$ (not a root).
   • $f(-1) = (-1)^3 + 4(-1)^2 + (-1) + 4 = -1 + 4 - 1 + 4 = 6$ (not a root).
   • $f(2) = 2^3 + 4(2)^2 + 2 + 4 = 8 + 16 + 2 + 4 = 30$ (not a root).
   • $f(-2) = (-2)^3 + 4(-2)^2 + (-2) + 4 = -8 + 16 - 2 + 4 = 10$ (not a root).
   • $f(4) = 4^3 + 4(4)^2 + 4 + 4 = 64 + 64 + 4 + 4 = 136$ (not a root).
   • $f(-4) = (-4)^3 + 4(-4)^2 + (-4) + 4 = -64 + 64 - 4 + 4 = 0$.

   $x = -4$ is a root.

3. Factor the polynomial: Now that we know $x = -4$ is a root, we can use synthetic division to factor $f(x)$ by $x + 4$.

   Dividing $x^3 + 4x^2 + x + 4$ by $x + 4$ yields: $x^3 + 4x^2 + x + 4 = (x + 4)(x^2 + 1)$

4. Solve for remaining factors: The remaining quadratic $x^2 + 1 = 0$ can be solved by setting it equal to zero: $x^2 + 1 = 0 \implies x^2 = -1 \implies x = \pm i$

Thus, the complete set of zeros for the polynomial is: $x = -4, i, -i$

Final Answer:

The zeros of $f(x) = x^3 + 4x^2 + x + 4$ are $-4, i, -i$.

Would you like more details on any part of this solution? Here are five related questions you can explore:

1. How does the Rational Root Theorem help in finding the zeros of a polynomial?
2. What is synthetic division, and how is it applied in factoring polynomials?
3. How do complex numbers arise in solving quadratic equations?
4. Can all cubic polynomials be factored into linear and quadratic terms?
5. What are the implications of having complex roots for the graph of the polynomial?

Tip: When solving polynomials, always test simple values first (like $x = 1, -1$) as they might help quickly identify potential roots.