To determine the full solution set for the equation \(x^3 + x^2 - 16x - 16 = 0\) given that \(x = -4\) is one solution, we can start by performing polynomial long division to factor the cubic polynomial.
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Dividing the polynomial by \(x + 4\) (since \(-4\) is a root):
- Divide \(x^3 + x^2 - 16x - 16\) by \(x + 4\).
Doing the division: \[ \begin{array}{r|rr} -4 & 1 & 1 & -16 & -16 \ & & -4 & 12 & 16 \ \hline & 1 & -3 & -4 & 0 \ \end{array} \] This gives us \(x^2 - 3x - 4\) after division.
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Factoring the quadratic: Now, we will factor the quadratic \(x^2 - 3x - 4\). We can look for two numbers that multiply to \(-4\) and add to \(-3\). These numbers are \(-4\) and \(1\). Thus, we can factor it as: \[ (x - 4)(x + 1) = 0 \]
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Setting the factors equal to zero: The complete factorization of the polynomial is: \[ (x + 4)(x - 4)(x + 1) = 0 \] Setting each factor to zero gives: \[ x + 4 = 0 \quad \Rightarrow \quad x = -4 \] \[ x - 4 = 0 \quad \Rightarrow \quad x = 4 \] \[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \]
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Conclusion: The full solution set is \(x = -4, 4, -1\). All these solutions are real numbers.
Thus, the correct statement describing the solution set is: The equation has all real solutions.