Find the value of x such that `(x+3)(3x-2)^(5)(7-x)^(3)(5x + 8)^(2) ge 0`

Options

A. `(-oo, -3] cup[2//3, 7]`
B. `(-oo, -3] cup [2//3, 7] cup {-8//5}`
C. `(-oo, -3] cup [2//3 - 8//5)`
D. None of these

Answer

To solve the inequality \((x+3)(3x-2)^{5}(7-x)^{3}(5x + 8)^{2} \geq 0\), we will follow these steps:

### Step 1: Identify the critical points
The critical points occur when each factor is equal to zero. We will set each factor to zero and solve for \(x\).

1. **From \(x + 3 = 0\)**:
\[ x = -3 \]

2. **From \(3x - 2 = 0\)**:
\[ 3x = 2 \implies x = \frac{2}{3} \]

3. **From \(7 - x = 0\)**:
\[ x = 7 \]

4. **From \(5x + 8 = 0\)**:
\[ 5x = -8 \implies x = -\frac{8}{5} \]

### Step 2: List the critical points
The critical points we found are:
- \(x = -3\)
- \(x = -\frac{8}{5}\) (which is approximately -1.6)
- \(x = \frac{2}{3}\)
- \(x = 7\)

### Step 3: Determine the intervals
The critical points divide the number line into the following intervals:
1. \((-∞, -3)\)
2. \((-3, -\frac{8}{5})\)
3. \((- \frac{8}{5}, \frac{2}{3})\)
4. \((\frac{2}{3}, 7)\)
5. \((7, ∞)\)

### Step 4: Test each interval
We will choose a test point from each interval to determine if the product is positive or negative.

1. **Interval \((-∞, -3)\)**: Choose \(x = -4\)
\[ (-4 + 3)(3(-4) - 2)^{5}(7 - (-4))^{3}(5(-4) + 8)^{2} = (-1)(-14)^{5}(11)^{3}(-12)^{2} \]
This is negative.

2. **Interval \((-3, -\frac{8}{5})\)**: Choose \(x = -2\)
\[ (-2 + 3)(3(-2) - 2)^{5}(7 - (-2))^{3}(5(-2) + 8)^{2} = (1)(-8)^{5}(9)^{3}( -2)^{2} \]
This is negative.

3. **Interval \((- \frac{8}{5}, \frac{2}{3})\)**: Choose \(x = 0\)
\[ (0 + 3)(3(0) - 2)^{5}(7 - 0)^{3}(5(0) + 8)^{2} = (3)(-2)^{5}(7)^{3}(8)^{2} \]
This is negative.

4. **Interval \((\frac{2}{3}, 7)\)**: Choose \(x = 1\)
\[ (1 + 3)(3(1) - 2)^{5}(7 - 1)^{3}(5(1) + 8)^{2} = (4)(1)^{5}(6)^{3}(13)^{2} \]
This is positive.

5. **Interval \((7, ∞)\)**: Choose \(x = 8\)
\[ (8 + 3)(3(8) - 2)^{5}(7 - 8)^{3}(5(8) + 8)^{2} = (11)(22)^{5}(-1)^{3}(48)^{2} \]
This is negative.

### Step 5: Compile the results
The function is non-negative in the intervals:
- At \(x = -3\) (where the function equals 0)
- In the interval \((\frac{2}{3}, 7)\) (where the function is positive)
- At \(x = \frac{2}{3}\) and \(x = 7\) (where the function equals 0)

### Final Answer
The solution to the inequality is:
\[ (-\infty, -3] \cup [\frac{2}{3}, 7] \]