Solve `((2x+3)(4-3x)^3(x-4))/((x-2)^2x^5)le0 `

Explanation

To solve the inequality \(\frac{(2x+3)(4-3x)^3(x-4)}{(x-2)^2x^5} \leq 0\), we will follow these steps:

### Step 1: Identify the critical points
The critical points occur where the numerator is zero or the denominator is zero.

1. **Numerator**: Set each factor to zero:
- \(2x + 3 = 0 \Rightarrow x = -\frac{3}{2}\)
- \(4 - 3x = 0 \Rightarrow x = \frac{4}{3}\)
- \(x - 4 = 0 \Rightarrow x = 4\)

2. **Denominator**: Set each factor to zero:
- \((x - 2)^2 = 0 \Rightarrow x = 2\)
- \(x^5 = 0 \Rightarrow x = 0\)

Thus, the critical points are:
- From the numerator: \(x = -\frac{3}{2}, \frac{4}{3}, 4\)
- From the denominator: \(x = 0, 2\)

### Step 2: Create a number line
Plot the critical points on a number line:
\[ -\infty, -\frac{3}{2}, 0, \frac{4}{3}, 2, 4, +\infty \]

### Step 3: Test intervals
We will test each interval defined by the critical points to determine where the function is negative or zero.

1. **Interval \((- \infty, -\frac{3}{2})\)**: Choose \(x = -2\)
\[ \frac{(2(-2)+3)(4-3(-2))^3(-2-4)}{((-2)-2)^2(-2)^5} = \frac{(-4+3)(4+6)^3(-6)}{(4)(-32)} = \frac{(-1)(10^3)(-6)}{(4)(-32)} > 0 \]

2. **Interval \((- \frac{3}{2}, 0)\)**: Choose \(x = -1\)
\[ \frac{(2(-1)+3)(4-3(-1))^3(-1-4)}{((-1)-2)^2(-1)^5} = \frac{(1)(7^3)(-5)}{(9)(-1)} < 0 \]

3. **Interval \((0, \frac{4}{3})\)**: Choose \(x = 1\)
\[ \frac{(2(1)+3)(4-3(1))^3(1-4)}{((1)-2)^2(1)^5} = \frac{(5)(1^3)(-3)}{(1)(1)} < 0 \]

4. **Interval \((\frac{4}{3}, 2)\)**: Choose \(x = 1.5\)
\[ \frac{(2(1.5)+3)(4-3(1.5))^3(1.5-4)}{((1.5)-2)^2(1.5)^5} = \frac{(6)(-0.5^3)(-2.5)}{(0.25)(7.59)} > 0 \]

5. **Interval \((2, 4)\)**: Choose \(x = 3\)
\[ \frac{(2(3)+3)(4-3(3))^3(3-4)}{((3)-2)^2(3)^5} = \frac{(9)(-5^3)(-1)}{(1)(243)} < 0 \]

6. **Interval \((4, +\infty)\)**: Choose \(x = 5\)
\[ \frac{(2(5)+3)(4-3(5))^3(5-4)}{((5)-2)^2(5)^5} = \frac{(13)(-11^3)(1)}{(9)(3125)} < 0 \]

### Step 4: Combine results
From our tests, we find:
- The function is negative or zero in the intervals:
- \((- \frac{3}{2}, 0)\)
- \((0, \frac{4}{3})\)
- \((2, 4)\)
- \((4, +\infty)\)

### Step 5: Include critical points
We include points where the function is zero:
- \(x = -\frac{3}{2}, \frac{4}{3}, 4\) (from the numerator)
- Exclude \(x = 0, 2\) (from the denominator)

### Final Solution
The solution set for the inequality \(\frac{(2x+3)(4-3x)^3(x-4)}{(x-2)^2x^5} \leq 0\) is:
\[ x \in \left(-\frac{3}{2}, 0\right) \cup \left(0, \frac{4}{3}\right) \cup (2, 4] \cup (4, +\infty) \]