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 one real solution and two non-real solutions.
The equation has one real solution and two non-real solutions.
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 two real solutions and one non-real solution.
The equation has two real solutions and one non-real solution.
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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 one real solution and two non-real solutions.
The equation has one real solution and two non-real solutions.
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 two real solutions and one non-real solution.
The equation has two real solutions and one non-real solution.
Skip to navigation
Answers
GPT-4o mini
To determine the full solution set of the equation \( x^3 + x^2 - 16x - 16 = 0 \) given that \( x = -4 \) is one of its solutions, we first need to perform polynomial division to factor the cubic equation.
Since \( x = -4 \) is a root, we can use synthetic division (or polynomial long division) to divide the cubic polynomial by \( x + 4 \).
### Synthetic Division
Using synthetic division with the root \( -4 \):
```
-4 | 1 1 -16 -16
| -4 12 16
----------------------
1 -3 -4 0
```
The quotient obtained is:
\[ x^2 - 3x - 4 \]
So we can write the original equation as:
\[ x^3 + x^2 - 16x - 16 = (x + 4)(x^2 - 3x - 4) \]
Next, we need to find the roots of the quadratic \( x^2 - 3x - 4 \). We can use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 1 \), \( b = -3 \), and \( c = -4 \):
\[
x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1}
\]
\[
x = \frac{3 \pm \sqrt{9 + 16}}{2}
\]
\[
x = \frac{3 \pm \sqrt{25}}{2}
\]
\[
x = \frac{3 \pm 5}{2}
\]
Calculating the two possible values:
1. \( x = \frac{8}{2} = 4 \)
2. \( x = \frac{-2}{2} = -1 \)
### Summary of Solutions
The complete solution set for the original equation \( x^3 + x^2 - 16x - 16 = 0 \) includes the roots:
- \( x = -4 \)
- \( x = 4 \)
- \( x = -1 \)
All of these solutions are real numbers.
### Final Conclusion
The correct statement regarding the solution set is:
**The equation has all real solutions.**
Since \( x = -4 \) is a root, we can use synthetic division (or polynomial long division) to divide the cubic polynomial by \( x + 4 \).
### Synthetic Division
Using synthetic division with the root \( -4 \):
```
-4 | 1 1 -16 -16
| -4 12 16
----------------------
1 -3 -4 0
```
The quotient obtained is:
\[ x^2 - 3x - 4 \]
So we can write the original equation as:
\[ x^3 + x^2 - 16x - 16 = (x + 4)(x^2 - 3x - 4) \]
Next, we need to find the roots of the quadratic \( x^2 - 3x - 4 \). We can use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 1 \), \( b = -3 \), and \( c = -4 \):
\[
x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1}
\]
\[
x = \frac{3 \pm \sqrt{9 + 16}}{2}
\]
\[
x = \frac{3 \pm \sqrt{25}}{2}
\]
\[
x = \frac{3 \pm 5}{2}
\]
Calculating the two possible values:
1. \( x = \frac{8}{2} = 4 \)
2. \( x = \frac{-2}{2} = -1 \)
### Summary of Solutions
The complete solution set for the original equation \( x^3 + x^2 - 16x - 16 = 0 \) includes the roots:
- \( x = -4 \)
- \( x = 4 \)
- \( x = -1 \)
All of these solutions are real numbers.
### Final Conclusion
The correct statement regarding the solution set is:
**The equation has all real solutions.**