Solve the following inequalities:
`2x^(3)-5x^(2)+2xle0`

Answer

To solve the inequality \( 2x^3 - 5x^2 + 2x \leq 0 \), we will follow these steps:

### Step 1: Factor the inequality
First, we can factor the expression on the left side of the inequality:

\[ 2x^3 - 5x^2 + 2x = x(2x^2 - 5x + 2) \]

Now, we need to solve the inequality:

\[ x(2x^2 - 5x + 2) \leq 0 \]

### Step 2: Factor the quadratic
Next, we will factor the quadratic \( 2x^2 - 5x + 2 \). We can use the method of splitting the middle term:

\[ 2x^2 - 4x - x + 2 = 0 \]

This can be factored as:

\[ (2x - 1)(x - 2) = 0 \]

Thus, we can rewrite the original inequality as:

\[ x(2x - 1)(x - 2) \leq 0 \]

### Step 3: Find the critical points
Now, we need to find the critical points where the expression equals zero:

1. \( x = 0 \)
2. \( 2x - 1 = 0 \) → \( x = \frac{1}{2} \)
3. \( x - 2 = 0 \) → \( x = 2 \)

So the critical points are \( x = 0, \frac{1}{2}, 2 \).

### Step 4: Test intervals
We will test the sign of the expression in the intervals determined by these critical points:

1. \( (-\infty, 0) \)
2. \( (0, \frac{1}{2}) \)
3. \( (\frac{1}{2}, 2) \)
4. \( (2, \infty) \)

**Interval 1: \( (-\infty, 0) \)**
Choose \( x = -1 \):
\[ (-1)(2(-1) - 1)(-1 - 2) = (-1)(-2 - 1)(-3) = (-1)(-3)(-3) = -9 \quad (\text{negative}) \]

**Interval 2: \( (0, \frac{1}{2}) \)**
Choose \( x = \frac{1}{4} \):
\[ \left(\frac{1}{4}\right)\left(2\left(\frac{1}{4}\right) - 1\right)\left(\frac{1}{4} - 2\right) = \left(\frac{1}{4}\right)\left(\frac{1}{2} - 1\right)\left(-\frac{7}{4}\right) = \left(\frac{1}{4}\right)(-\frac{1}{2})(-\frac{7}{4}) = \frac{7}{32} \quad (\text{positive}) \]

**Interval 3: \( (\frac{1}{2}, 2) \)**
Choose \( x = 1 \):
\[ (1)(2(1) - 1)(1 - 2) = (1)(2 - 1)(-1) = (1)(1)(-1) = -1 \quad (\text{negative}) \]

**Interval 4: \( (2, \infty) \)**
Choose \( x = 3 \):
\[ (3)(2(3) - 1)(3 - 2) = (3)(6 - 1)(1) = (3)(5)(1) = 15 \quad (\text{positive}) \]

### Step 5: Determine the solution
From our tests, we have the following signs for the intervals:

1. \( (-\infty, 0) \): Negative
2. \( (0, \frac{1}{2}) \): Positive
3. \( (\frac{1}{2}, 2) \): Negative
4. \( (2, \infty) \): Positive

Since we are looking for where the expression is less than or equal to zero, we include the intervals where the expression is negative and the points where it equals zero:

- The solution is \( (-\infty, 0] \cup [\frac{1}{2}, 2] \).

### Final Answer:
\[ x \in (-\infty, 0] \cup [\frac{1}{2}, 2] \]