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 non-real solutions.%0D%0AThe equation has all non-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 real solutions.%0D%0AThe equation has all real solutions.%0D%0A%0D%0AThe equation has one real solution and two non-real solutions.

Answer 1

To determine the full solution set for the equation \(x^3 + x^2 - 16x - 16 = 0\), we start by verifying the given root \(-4\).

**Step 1: Verify if \(-4\) is a root.**

Substituting \(x = -4\) into the equation:
\[
(-4)^3 + (-4)^2 - 16(-4) - 16 = -64 + 16 + 64 - 16 = 0.
\]
Since the expression equals \(0\), \(-4\) is indeed a root.

**Step 2: Factor the polynomial.**

Since \(-4\) is a root, we can factor the polynomial using synthetic division or polynomial long division. Let's perform synthetic division with \(-4\):

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

The result shows that:
\[
x^3 + x^2 - 16x - 16 = (x + 4)(x^2 - 3x - 4).
\]

**Step 3: Factor the quadratic.**

Now, we need to factor the quadratic \(x^2 - 3x - 4\). We can either use the quadratic formula or factor it directly:
\[
x^2 - 3x - 4 = (x - 4)(x + 1).
\]

**Step 4: Write the full factorization.**

The complete factorization of the original polynomial is:
\[
(x + 4)(x - 4)(x + 1) = 0.
\]

**Step 5: Find the roots.**

Setting each factor to zero gives the roots:
1. \(x + 4 = 0 \Rightarrow x = -4\)
2. \(x - 4 = 0 \Rightarrow x = 4\)
3. \(x + 1 = 0 \Rightarrow x = -1\)

The solutions to the equation are:
\[
x = -4, \quad x = 4, \quad x = -1.
\]

**Conclusion:**

Since all solutions are real numbers, the statement that accurately describes the solution set is:

**The equation has all real solutions.**