Ask a New Question
Menu

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 real solutions.
The equation has all real solutions.

The equation has all non-real solutions.
The equation has all non-real solutions.

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.

1 answer

To solve the equation \(x^3 + x^2 - 16x - 16 = 0\) and confirm that \(-4\) is indeed a solution, we can perform polynomial division.

First, we can factor the polynomial using the known root \(-4\):

  1. Since \(-4\) is a root, we can use synthetic division to divide the polynomial by \((x + 4)\).

The coefficients of the polynomial \(x^3 + x^2 - 16x - 16\) are \(1, 1, -16, -16\).

Performing synthetic division:

-4 | 1   1   -16   -16
   |       -4      12    16
-------------------------
     1  -3     -4      0

The result is \(x^2 - 3x - 4\) with a remainder of \(0\), confirming that:

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

  1. Now we factor the quadratic \(x^2 - 3x - 4\):

To factor \(x^2 - 3x - 4\), we look for two numbers that multiply to \(-4\) and add to \(-3\). These numbers are \(-4\) and \(1\), so we can factor it as:

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

  1. Now, we can write the fully factored form of the original polynomial:

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

  1. To find the solutions to the equation, we set each factor equal to zero:
  • \(x + 4 = 0 \Rightarrow x = -4\)
  • \(x - 4 = 0 \Rightarrow x = 4\)
  • \(x + 1 = 0 \Rightarrow x = -1\)

Thus, the complete solution set is \({-4, 4, -1}\), and all of these solutions are indeed real numbers.

Conclusion: The correct response is:

The equation has all real solutions.