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

Answer 1

To solve the equation \( x^3 + x^2 - 16x - 16 = 0 \) and verify that \( x = -4 \) is indeed a solution, we first perform synthetic division to factor the polynomial using this root.

### Step 1: Synthetic Division

Using synthetic division with \( x = -4 \):

```
-4 | 1 1 -16 -16
| -4 12 16
------------------------
1 -3 -4 0
```

After performing synthetic division, we find:

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

### Step 2: Factor the Quadratic

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

We look for two numbers that multiply to \(-4\) and add to \(-3\). The numbers \(-4\) and \(1\) fit this criteria. Thus, we can factor the quadratic as follows:

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

### Step 3: Write the Full Factorization

Thus, the original polynomial can be factored as:

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

### Step 4: Find the Roots

Now, we can set each factor equal to zero to find the solutions:

1. \( x + 4 = 0 \) → \( x = -4 \)
2. \( x - 4 = 0 \) → \( x = 4 \)
3. \( x + 1 = 0 \) → \( x = -1 \)

### Step 5: Solution Set

The solutions to the equation \( x^3 + x^2 - 16x - 16 = 0 \) are:

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

### Conclusion

All of the solutions are real. Therefore, the correct statement describing the solution set is:

**The equation has all real solutions.**