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The expression you've provided is:

\[
\frac{x + 9}{\sqrt{-4x - 6}}
\]

To factor this expression, let's consider both the numerator and the expression inside the square root separately.

1. **Numerator: \( x + 9 \)**  
   This is a linear expression, and it is already in its simplest form.

2. **Denominator: \( \sqrt{-4x - 6} \)**  
   First, let's factor the expression inside the square root.

   \[
   -4x - 6 = -2(2x + 3)
   \]

   Therefore, the expression inside the square root can be rewritten as:

   \[
   \sqrt{-4x - 6} = \sqrt{-2(2x + 3)}
   \]

   Now, the entire expression becomes:

   \[
   \frac{x + 9}{\sqrt{-2(2x + 3)}}
   \]

Since the expression inside the square root is already simplified, the final factored form of the expression is:

\[
\frac{x + 9}{\sqrt{-2} \cdot \sqrt{2x + 3}} = \frac{x + 9}{\sqrt{2} \cdot \sqrt{-1} \cdot \sqrt{2x + 3}}
\]

Given that \( \sqrt{-1} \) is the imaginary unit \( i \), the expression can further be written as:

\[
\frac{x + 9}{i\sqrt{2} \cdot \sqrt{2x + 3}} = \frac{x + 9}{\sqrt{2} \cdot i \cdot \sqrt{2x + 3}}
\]

However, the expression doesn't factor in any simpler form unless considering the imaginary unit \( i \) as a factor.

Would you like to explore any other aspect of this problem? 

Here are 5 related questions you might find useful:
1. How can you find the domain of the given expression?
2. How would you simplify the expression if it had real values only?
3. What happens if the denominator is set to zero?
4. How would you differentiate this expression with respect to \(x\)?
5. Can you solve the equation \( x + 9 = 0 \) for \( x \)?

**Tip:** When factoring expressions with square roots, consider if the radicand (the expression inside the square root) can be simplified by factoring out constants or perfect squares.

Explore how to factorize the complex expression (x + 9) / sqrt(-4x - 6) with step-by-step guidance. This problem involves algebraic manipulation, including the quotient rule for square roots and understanding complex numbers with the imaginary unit (i). Suitable for advanced high school students.

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