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:
- \( x = \frac{8}{2} = 4 \)
- \( x = \frac{-2}{2} = -1 \)
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.
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