To solve the inequality \(-x^3 - 6x^2 - 9x \leq 0\), we start by rewriting it:
\[ -x^3 - 6x^2 - 9x \leq 0 \]
We can factor out a negative sign from the inequality:
\[ -(x^3 + 6x^2 + 9x) \leq 0 \]
This simplifies to:
\[ x^3 + 6x^2 + 9x \geq 0 \]
Next, we factor the cubic polynomial. First, we can factor out an \(x\):
\[ x(x^2 + 6x + 9) \geq 0 \]
Now, the quadratic \(x^2 + 6x + 9\) can be factored further:
\[ x^2 + 6x + 9 = (x + 3)^2 \]
So, we can rewrite our inequality as:
\[ x(x + 3)^2 \geq 0 \]
Next, we find the critical points by setting each factor equal to zero:
- \(x = 0\)
- \((x + 3)^2 = 0 \Rightarrow x = -3\)
The critical points are \(x = 0\) and \(x = -3\). Now, we will test the intervals defined by these critical points: \( (-\infty, -3) \), \( (-3, 0) \), and \( (0, \infty) \).
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Interval \( (-\infty, -3) \):
- Choose \(x = -4\): \[ (-4)((-4) + 3)^2 = (-4)(-1)^2 = -4 < 0 \quad \text{(not a solution)} \]
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Interval \( (-3, 0) \):
- Choose \(x = -2\): \[ (-2)((-2) + 3)^2 = (-2)(1)^2 = -2 < 0 \quad \text{(not a solution)} \]
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Interval \( (0, \infty) \):
- Choose \(x = 1\): \[ (1)((1) + 3)^2 = (1)(4)^2 = 16 > 0 \quad \text{(solution)} \]
We also need to check the endpoints:
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At \(x = -3\): \[ (-3)((-3) + 3)^2 = (-3)(0)^2 = 0 \quad \text{(included)} \]
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At \(x = 0\): \[ (0)((0) + 3)^2 = (0)(3)^2 = 0 \quad \text{(included)} \]
From this analysis, the solution set is:
- \(x \in [0, \infty)\)
- \(x = -3\)
In interval notation, the solution is:
\[ [-3, 0) \cup [0, \infty) \]
So the best answer matching the options provided is A. The solutions are the values of \(x\) in the interval \([0, \infty)\) as well as the point \(x = -3\).
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