Solve `((x-3)(x+5)(x-7))/(|x-4|(x+6))le0`

Explanation

To solve the inequality

\[ \frac{(x-3)(x+5)(x-7)}{|x-4|(x+6)} \leq 0, \]

we will follow these steps:

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

- **Numerator**: Set \((x-3)(x+5)(x-7) = 0\).
- \(x - 3 = 0 \Rightarrow x = 3\)
- \(x + 5 = 0 \Rightarrow x = -5\)
- \(x - 7 = 0 \Rightarrow x = 7\)

- **Denominator**: Set \(|x-4|(x+6) = 0\).
- \(|x-4| = 0 \Rightarrow x = 4\) (undefined at this point)
- \(x + 6 = 0 \Rightarrow x = -6\) (undefined at this point)

So, the critical points are: \(x = -6, -5, 3, 4, 7\).

### Step 2: Create a number line
Plot the critical points on a number line:

\[ -\infty \quad -6 \quad -5 \quad 3 \quad 4 \quad 7 \quad +\infty \]

### Step 3: Test intervals
We will test the sign of the expression in each interval determined by the critical points:

1. **Interval \((- \infty, -6)\)**: Choose \(x = -7\)
\[ \frac{(-7-3)(-7+5)(-7-7)}{|(-7-4)|(-7+6)} = \frac{(-10)(-2)(-14)}{11(-1)} < 0 \]

2. **Interval \((-6, -5)\)**: Choose \(x = -5.5\)
\[ \frac{(-5.5-3)(-5.5+5)(-5.5-7)}{|(-5.5-4)|(-5.5+6)} = \frac{(-8.5)(-0.5)(-12.5)}{9.5(0.5)} < 0 \]

3. **Interval \((-5, 3)\)**: Choose \(x = 0\)
\[ \frac{(0-3)(0+5)(0-7)}{|(0-4)|(0+6)} = \frac{(-3)(5)(-7)}{4(6)} > 0 \]

4. **Interval \((3, 4)\)**: Choose \(x = 3.5\)
\[ \frac{(3.5-3)(3.5+5)(3.5-7)}{|(3.5-4)|(3.5+6)} = \frac{(0.5)(8.5)(-3.5)}{0.5(9.5)} < 0 \]

5. **Interval \((4, 7)\)**: Choose \(x = 5\)
\[ \frac{(5-3)(5+5)(5-7)}{|(5-4)|(5+6)} = \frac{(2)(10)(-2)}{1(11)} < 0 \]

6. **Interval \((7, +\infty)\)**: Choose \(x = 8\)
\[ \frac{(8-3)(8+5)(8-7)}{|(8-4)|(8+6)} = \frac{(5)(13)(1)}{4(14)} > 0 \]

### Step 4: Compile results
From the tests, we find:
- Negative in intervals: \((- \infty, -6)\), \((-6, -5)\), \((3, 4)\), and \((4, 7)\).
- The expression is zero at \(x = -5, 3, 7\).

### Step 5: Write the solution
The solution to the inequality \(\frac{(x-3)(x+5)(x-7)}{|x-4|(x+6)} \leq 0\) is:

\[ x \in (-\infty, -6) \cup (-6, -5] \cup [3, 4) \cup (4, 7]. \]