Question

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 non-real solutions.
The equation has all non-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 one real solution and two non-real solutions.
The equation has one real solution and two non-real solutions.

The equation has all real solutions.
The equation has all real solutions.

Answer 1

To analyze the given polynomial equation \( x^3 + x^2 - 16x - 16 = 0 \) with the known solution \( x = -4 \), we can perform synthetic division to find the other roots.

1. **Synthetic Division**:
We divide \( x^3 + x^2 - 16x - 16 \) by \( x + 4 \):

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

This shows that:

\[
x^3 + x^2 - 16x - 16 = (x + 4)(x^2 - 3x - 4)
\]

2. **Factoring the Quadratic**:
Next, we factor the quadratic \( x^2 - 3x - 4 \). We find two numbers that multiply to -4 and add to -3: those numbers are -4 and 1.

Thus, we can write:

\[
x^2 - 3x - 4 = (x - 4)(x + 1)
\]

3. **Complete Factorization**:
Putting it all together, we have:

\[
x^3 + x^2 - 16x - 16 = (x + 4)(x - 4)(x + 1)
\]

4. **Finding the Roots**:
The solutions to the equation are:

\[
x + 4 = 0 \implies x = -4
\]
\[
x - 4 = 0 \implies x = 4
\]
\[
x + 1 = 0 \implies x = -1
\]

Thus, the full solution set is \( x = -4, 4, -1 \), and all of these are real solutions.

5. **Conclusion**:
The statement that correctly describes the solution set is:

**The equation has all real solutions.**