One of the solutions to the equation x3+x2−16x−16=0 is −4 .

To solve the equation \(x^3 + x^2 - 16x - 16 = 0\) with one known solution \(x = -4\), we can perform synthetic division or polynomial long division to divide the cubic polynomial by \((x + 4)\), since \(-4\) is a root.

### Synthetic Division

We can set up synthetic division:

- Coefficients of the polynomial: \(1\) (for \(x^3\)), \(1\) (for \(x^2\)), \(-16\) (for \(x\)), and \(-16\) (constant term).
- We divide by \(-4\):

\[
\begin{array}{r|rrrr}
-4 & 1 & 1 & -16 & -16 \\
& & -4 & 12 & 16 \\
\hline
& 1 & -3 & -4 & 0 \\
\end{array}
\]

The result is:

\[
x^2 - 3x - 4
\]

Now we can factor \(x^2 - 3x - 4\):

### Factoring the Quadratic

We look for two numbers that multiply to \(-4\) and add to \(-3\). These numbers are \(-4\) and \(1\).

Thus, we can factor the quadratic as:

\[
(x - 4)(x + 1) = 0
\]

### Solving for Roots

Now we have:

1. The first solution from the linear term: \(x + 4 = 0 \Rightarrow x = -4\)
2. The second solution from \(x - 4 = 0 \Rightarrow x = 4\)
3. The third solution from \(x + 1 = 0 \Rightarrow x = -1\)

### Complete Solution Set

The full solution set for the equation \(x^3 + x^2 - 16x - 16 = 0\) is:

\[
x = -4, \, x = 4, \, x = -1
\]

### Conclusion

All the solutions are real.

Thus, the correct statement about the solution set is:

**The equation has all real solutions.**