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.

1 answer

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:

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

  1. 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:

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

  1. Complete Factorization: Putting it all together, we have:

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

  1. 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.

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

The equation has all real solutions.