To determine the full solution set for the equation \(x^3 + x^2 - 16x - 16 = 0\) given that \(x = -4\) is one solution, we can start by performing polynomial long division to factor the cubic polynomial.
1. **Dividing the polynomial by \(x + 4\)** (since \(-4\) is a root):
- Divide \(x^3 + x^2 - 16x - 16\) by \(x + 4\).
Doing the division:
\[
\begin{array}{r|rr}
-4 & 1 & 1 & -16 & -16 \\
& & -4 & 12 & 16 \\
\hline
& 1 & -3 & -4 & 0 \\
\end{array}
\]
This gives us \(x^2 - 3x - 4\) after division.
2. **Factoring the quadratic**:
Now, we will factor the quadratic \(x^2 - 3x - 4\).
We can look for two numbers that multiply to \(-4\) and add to \(-3\). These numbers are \(-4\) and \(1\).
Thus, we can factor it as:
\[
(x - 4)(x + 1) = 0
\]
3. **Setting the factors equal to zero**:
The complete factorization of the polynomial is:
\[
(x + 4)(x - 4)(x + 1) = 0
\]
Setting each factor to zero gives:
\[
x + 4 = 0 \quad \Rightarrow \quad x = -4
\]
\[
x - 4 = 0 \quad \Rightarrow \quad x = 4
\]
\[
x + 1 = 0 \quad \Rightarrow \quad x = -1
\]
4. **Conclusion**:
The full solution set is \(x = -4, 4, -1\). All these solutions are real numbers.
Thus, the correct statement describing the solution set is:
**The equation has all real solutions.**