One of the solutions to the equation x3+x2−16x−16=0 is −4 . Test the solution to find the full solution set. Which of the following statements correctly describes the solution set? (1 point) Responses The equation has all real solutions. The equation has all real solutions. The equation has two real solutions and one non-real solution. The equation has two real solutions and one non-real solution. The equation has all non-real solutions. The equation has all non-real solutions. The equation has one real solution and two non-real solutions

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.**