The problem given is:

$f(x) = x^3 + x^2 - 14x - 24$

You are asked to find all the zeros of $f(x)$, knowing that one of the zeros is $x = 4$.

Step 1: Use Synthetic Division or Polynomial Division

Since $x = 4$ is a zero of the polynomial, we can divide the polynomial $f(x)$ by $(x - 4)$.

Using synthetic division to divide $f(x)$ by $(x - 4)$:

1. Write down the coefficients of the polynomial: $1, 1, -14, -24$.
2. Perform synthetic division using $4$:

   4 |   1    1    -14   -24
      |        4    20    24
      ----------------------
        1    5     6     0

The quotient is $x^2 + 5x + 6$, and the remainder is 0 (which confirms $x = 4$ is indeed a root).

Step 2: Factor the Quotient

Now, we need to solve $x^2 + 5x + 6 = 0$.

Factorizing the quadratic: $x^2 + 5x + 6 = (x + 2)(x + 3)$

Step 3: Solve for the Remaining Zeros

From the factorization, we get: $x + 2 = 0 \quad \text{or} \quad x + 3 = 0$ So, $x = -2$ and $x = -3$.

Final Answer:

The zeros of $f(x) = x^3 + x^2 - 14x - 24$ are: $x = 4, \quad x = -2, \quad x = -3$

Would you like further details on synthetic division or how the quadratic was factored?

Here are 5 related questions to deepen your understanding:

1. How do you perform synthetic division for higher-degree polynomials?
2. What are the steps to factor a quadratic polynomial?
3. Can you use the Rational Root Theorem to find potential zeros of a cubic polynomial?
4. How does synthetic division confirm that a number is a zero of a polynomial?
5. How would the approach change if a complex root was involved?

Tip: Always check your work by plugging the found roots back into the original polynomial to verify that they make the equation equal to zero.