Solve. $$ 4 x^{3}-5 x^{2} \leq 21 x $$ The solution set is $$ \square $$. (Type your answer in interval notation.)

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We start with the inequality
$$
4x^3 - 5x^2 \le 21x.
$$

**Step 1.** Move all terms to one side:
$$
4x^3 - 5x^2 - 21x \le 0.
$$

**Step 2.** Factor out the common term $$x$$:
$$
x\left(4x^2 - 5x - 21\right) \le 0.
$$

**Step 3.** Factor the quadratic $$4x^2 - 5x - 21$$. Compute the discriminant:
$$
\Delta = (-5)^2 - 4(4)(-21) = 25 + 336 = 361.
$$
Since $$\sqrt{361} = 19$$, the roots are
$$
x = \frac{5 \pm 19}{2 \cdot 4}.
$$
This gives
$$
x = \frac{24}{8} = 3 \quad \text{and} \quad x = \frac{-14}{8} = -\frac{7}{4}.
$$

The quadratic factors as
$$
4x^2 - 5x - 21 = 4\left(x - 3\right)\left(x + \frac{7}{4}\right).
$$

**Step 4.** Rewrite the inequality (dividing by the positive constant $$4$$ does not change the inequality):
$$
x \left(x - 3\right)\left(x + \frac{7}{4}\right) \le 0.
$$

**Step 5.** Identify the critical points:
$$
x = -\frac{7}{4}, \quad x = 0, \quad x = 3.
$$

These points divide the real number line into four intervals:
- $$(-\infty, -\frac{7}{4})$$
- $$(-\frac{7}{4}, 0)$$
- $$(0, 3)$$
- $$(3, \infty)$$

**Step 6.** Analyze the sign of the product $$x\left(x - 3\right)\left(x + \frac{7}{4}\right)$$:

1. For $$x < -\frac{7}{4}$$ (choose $$x = -2$$):
   - $$x$$ is negative.
   - $$x - 3$$ is negative.
   - $$x + \frac{7}{4}$$ is negative.
   
   The product: negative $$\times$$ negative $$\times$$ negative = negative (satisfies $$\le 0$$).

2. For $$-\frac{7}{4} < x < 0$$ (choose $$x = -1$$):
   - $$x$$ is negative.
   - $$x - 3$$ is negative.
   - $$x + \frac{7}{4}$$ is positive.
   
   The product: negative $$\times$$ negative $$\times$$ positive = positive (does not satisfy).

3. For $$0 < x < 3$$ (choose $$x = 1$$):
   - $$x$$ is positive.
   - $$x - 3$$ is negative.
   - $$x + \frac{7}{4}$$ is positive.
   
   The product: positive $$\times$$ negative $$\times$$ positive = negative (satisfies).

4. For $$x > 3$$ (choose $$x = 4$$):
   - $$x$$ is positive.
   - $$x - 3$$ is positive.
   - $$x + \frac{7}{4}$$ is positive.
   
   The product: positive $$\times$$ positive $$\times$$ positive = positive (does not satisfy).

**Step 7.** Include the points where the product is zero. These are $$x = -\frac{7}{4}$$, $$x = 0$$, and $$x = 3$$.

Thus, the solution set in interval notation is:
$$
(-\infty, -\tfrac{7}{4}] \cup [0, 3].
$$
The solution set is $$ (-\infty, -\frac{7}{4}] \cup [0, 3] $$.

Solve.

The solution set is . (Type your answer in interval notation.)

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