To confirm that \( x = -4 \) is a solution to the equation \( x^3 + x^2 - 16x - 16 = 0 \), we substitute it into the equation:
\[ (-4)^3 + (-4)^2 - 16(-4) - 16 \] \[ = -64 + 16 + 64 - 16 \] \[ = -64 + 16 + 64 - 16 = 0 \]
This verifies that \( x = -4 \) is indeed a solution.
Next, we can factor the polynomial \( x^3 + x^2 - 16x - 16 \) given that one solution is \( x + 4 \). We'll use synthetic or polynomial division, treating \( x + 4 \) as a known factor.
Performing synthetic division on \( x^3 + x^2 - 16x - 16 \) using \( x = -4 \):
- Write coefficients: \( 1, 1, -16, -16 \)
- Setup synthetic division with -4:
-4 | 1 1 -16 -16
| -4 12 16
----------------------
1 -3 -4 0
After synthetic division, we find that \( x^3 + x^2 - 16x - 16 \) can be factored as:
\[ (x + 4)(x^2 - 3x - 4) \]
Next, we need to solve the quadratic equation \( x^2 - 3x - 4 = 0 \) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \( a = 1, b = -3, c = -4 \):
\[ x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-4)}}{2(1)} \] \[ x = \frac{3 \pm \sqrt{9 + 16}}{2} \] \[ x = \frac{3 \pm \sqrt{25}}{2} \] \[ x = \frac{3 \pm 5}{2} \]
Calculating the two roots:
- \( x = \frac{8}{2} = 4 \)
- \( x = \frac{-2}{2} = -1 \)
Thus, the full solution set for the equation \( x^3 + x^2 - 16x - 16 = 0 \) is:
- \( x = -4 \)
- \( x = 4 \)
- \( x = -1 \)
All solutions are real.
Statement that correctly describes the solution set: The equation has all real solutions.