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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.

Find all the zeros of f(x) = x^3 + x^2 - 14x - 24 if one of the zeros is 4.

Learn how to find the zeros of the polynomial f(x) = x^3 + x^2 - 14x - 24 using synthetic division and quadratic factorization. This problem is solved step-by-step and is ideal for students in grades 10-12.

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