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:
- \(x + 4 = 0 \Rightarrow x = -4\)
- \(x - 4 = 0 \Rightarrow x = 4\)
- \(x + 1 = 0 \Rightarrow x = -1\)
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.
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