To find a polynomial of degree 3 with real coefficients and given conditions, we'll follow these steps:

Step 1: Start with the factored form

Since the polynomial has zeros at $x = -3$, $x = -1$, and $x = 4$, the factored form of the polynomial can be written as:

$f(x) = a(x + 3)(x + 1)(x - 4)$

Here, $a$ is a constant that we need to determine.

Step 2: Use the condition $f(-2) = 12$

We're given that when $x = -2$, the polynomial evaluates to 12, i.e., $f(-2) = 12$. We can substitute $x = -2$ into the polynomial and solve for $a$.

$f(-2) = a((-2 + 3)(-2 + 1)(-2 - 4)) = 12$

Simplify the terms:

$f(-2) = a(1)(-1)(-6) = 12$

$f(-2) = 6a = 12$

Step 3: Solve for $a$

Now, solve for $a$:

$a = \frac{12}{6} = 2$

Step 4: Write the polynomial

Now that we know $a = 2$, the polynomial becomes:

$f(x) = 2(x + 3)(x + 1)(x - 4)$

Step 5: Expand the polynomial

Let's expand the polynomial:

First, expand the first two factors: $(x + 3)(x + 1) = x^2 + 4x + 3$

Now, expand this result with $(x - 4)$: $(x^2 + 4x + 3)(x - 4) = x^3 - 4x^2 + 4x^2 - 16x + 3x - 12$ $= x^3 - 13x - 12$

Finally, multiply by 2: $f(x) = 2(x^3 - 13x - 12) = 2x^3 - 26x - 24$

Final answer:

The polynomial is: $f(x) = 2x^3 - 26x - 24$

Would you like more details on the steps, or do you have any questions?

Here are 5 related questions to explore further:

1. How do you find the degree of a polynomial?
2. Can you solve for $a$ in a different way?
3. What is the significance of the constant $a$ in polynomial equations?
4. How would you factor the expanded polynomial back to its roots?
5. What happens if a polynomial has complex zeros?

Tip: When solving polynomial equations, always check the conditions given (like function values at certain points) to adjust the equation properly.