One of the solutions to the equation x3+x2−16x−16=0 is −4 .

Question

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 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.

Answer 1

To solve the equation \(x^3 + x^2 - 16x - 16 = 0\) with one known solution \(x = -4\), we can perform synthetic division or polynomial long division to divide the cubic polynomial by \((x + 4)\), since \(-4\) is a root.

Synthetic Division

We can set up synthetic division:

• Coefficients of the polynomial: \(1\) (for \(x^3\)), \(1\) (for \(x^2\)), \(-16\) (for \(x\)), and \(-16\) (constant term).
• We divide by \(-4\):

\[ \begin{array}{r|rrrr} -4 & 1 & 1 & -16 & -16 \ & & -4 & 12 & 16 \ \hline & 1 & -3 & -4 & 0 \ \end{array} \]

The result is:

\[ x^2 - 3x - 4 \]

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

Factoring the Quadratic

We look for two numbers that multiply to \(-4\) and add to \(-3\). These numbers are \(-4\) and \(1\).

Thus, we can factor the quadratic as:

\[ (x - 4)(x + 1) = 0 \]

Solving for Roots

Now we have:

1. The first solution from the linear term: \(x + 4 = 0 \Rightarrow x = -4\)
2. The second solution from \(x - 4 = 0 \Rightarrow x = 4\)
3. The third solution from \(x + 1 = 0 \Rightarrow x = -1\)

Complete Solution Set

The full solution set for the equation \(x^3 + x^2 - 16x - 16 = 0\) is:

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

Conclusion

All the solutions are real.

Thus, the correct statement about the solution set is:

The equation has all real solutions.