Question

One of the solutions to the equation x3+x2−16x−16=0 is −4.%0D%0A%0D%0ATest the solution to find the full solution set.%0D%0A%0D%0AWhich of the following statements correctly describes the solution set?%0D%0A%0D%0A(1 point)%0D%0AResponses%0D%0A%0D%0AThe equation has all real solutions.%0D%0AThe equation has all real solutions.%0D%0A%0D%0AThe equation has two real solutions and one non-real solution.%0D%0AThe equation has two real solutions and one non-real solution.%0D%0A%0D%0AThe equation has all non-real solutions.%0D%0AThe equation has all non-real solutions.%0D%0A%0D%0AThe equation has one real solution and two non-real solutions.

Answer 1

To solve the equation \(x^3 + x^2 - 16x - 16 = 0\) and determine the nature of its solutions, we can start by verifying that \(-4\) is indeed a solution.

Substituting \(x = -4\) into the equation:

\[
(-4)^3 + (-4)^2 - 16(-4) - 16 = -64 + 16 + 64 - 16 = -64 + 16 + 64 - 16 = 0
\]

Since the equation evaluates to 0, \(-4\) is a solution.

Next, we can use polynomial long division to factor the cubic polynomial \(x^3 + x^2 - 16x - 16\) by dividing it by \(x + 4\) (which corresponds to the root \(-4\)).

1. **Divide** \(x^3 + x^2 - 16x - 16\) by \(x + 4\).

- \(x^3 \div x = x^2\)
- Multiply \(x^2\) by \(x + 4\): \(x^3 + 4x^2\)
- Subtract: \((x^3 + x^2) - (x^3 + 4x^2) = -3x^2 - 16x\)

- Next, divide \(-3x^2 \div x = -3x\)
- Multiply \(-3x\) by \(x + 4\): \(-3x^2 - 12x\)
- Subtract: \((-3x^2 - 16x) - (-3x^2 - 12x) = -4x - 16\)

- Finally, divide \(-4x \div x = -4\)
- Multiply \(-4\) by \(x + 4\): \(-4x - 16\)
- Subtract: \((-4x - 16) - (-4x - 16) = 0\)

Thus, we have:

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

Now, we need to solve the quadratic \(x^2 - 3x - 4 = 0\) using the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \(a = 1\), \(b = -3\), and \(c = -4\):

\[
x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1}
\]
\[
x = \frac{3 \pm \sqrt{9 + 16}}{2}
\]
\[
x = \frac{3 \pm \sqrt{25}}{2}
\]
\[
x = \frac{3 \pm 5}{2}
\]

Thus, we have two solutions:

\[
x = \frac{8}{2} = 4 \quad \text{and} \quad x = \frac{-2}{2} = -1
\]

So the complete set of solutions to \(x^3 + x^2 - 16x - 16 = 0\) is:

1. \( x = -4 \)
2. \( x = 4 \)
3. \( x = -1 \)

All solutions are real numbers.

Thus, the correct statement that describes the solution set is:

**The equation has all real solutions.**