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 \(-4\)).
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Divide \(x^3 + x^2 - 16x - 16\) by \(x + 4\).
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\(x^3 \div x = x^2\)
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Multiply \(x^2\) by \(x + 4\): \(x^3 + 4x^2\)
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Subtract: \((x^3 + x^2) - (x^3 + 4x^2) = -3x^2 - 16x\)
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Next, divide \(-3x^2 \div x = -3x\)
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Multiply \(-3x\) by \(x + 4\): \(-3x^2 - 12x\)
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Subtract: \((-3x^2 - 16x) - (-3x^2 - 12x) = -4x - 16\)
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Finally, divide \(-4x \div x = -4\)
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Multiply \(-4\) by \(x + 4\): \(-4x - 16\)
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Subtract: \((-4x - 16) - (-4x - 16) = 0\)
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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:
- \( x = -4 \)
- \( x = 4 \)
- \( x = -1 \)
All solutions are real numbers.
Thus, the correct statement that describes the solution set is:
The equation has all real solutions.