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The problem asks to find and simplify \(\frac{h(x)}{g(x)}\), where:

\[
h(x) = 4x^2 + 5x - 9
\]
and
\[
g(x) = x - 1
\]

### Step-by-Step Solution:

1. **Factor \( h(x) \)**:
   We need to factor \( h(x) = 4x^2 + 5x - 9 \). To factor this quadratic expression, we use the method of splitting the middle term or the quadratic formula.

   First, we find two numbers whose product is \( 4 \times (-9) = -36 \) and whose sum is \( 5 \). These numbers are \( 9 \) and \( -4 \).

   Now split the middle term:
   \[
   4x^2 + 9x - 4x - 9
   \]

   Group the terms:
   \[
   (4x^2 - 4x) + (9x - 9)
   \]

   Factor each group:
   \[
   4x(x - 1) + 9(x - 1)
   \]

   Factor out the common term \( (x - 1) \):
   \[
   (x - 1)(4x + 9)
   \]

   So, \( h(x) = (x - 1)(4x + 9) \).

2. **Simplify \( \frac{h(x)}{g(x)} \)**:
   Now, we can simplify the given expression:
   \[
   \frac{h(x)}{g(x)} = \frac{(x - 1)(4x + 9)}{x - 1}
   \]

   Since \( x - 1 \) is a common factor in both the numerator and the denominator, we can cancel it out, assuming \( x \neq 1 \):
   \[
   \frac{h(x)}{g(x)} = 4x + 9
   \]

### Final Answer:
\[
\frac{h(x)}{g(x)} = 4x + 9 \quad \text{for} \quad x \neq 1
\]

Let me know if you'd like more details on any step or have any further questions! Here are some related questions:

1. What happens to the expression if \( x = 1 \)?
2. How would the process differ if \( g(x) \) was more complex?
3. Can you always cancel terms like \( x - 1 \) in the numerator and denominator?
4. How do you handle cases where factoring isn't possible?
5. What are other factoring methods for quadratic equations?

**Tip**: Always check if factors can cancel, but ensure to exclude any values that would make the denominator zero!

Find and simplify h(x)/g(x), hint: you will need to factor h(x). h(x) = 4x^2 + 5x - 9 and g(x) = x - 1.

Learn how to simplify the rational expression h(x)/g(x), where h(x) = 4x^2 + 5x - 9 and g(x) = x - 1. This problem demonstrates factoring quadratic expressions and simplifying rational expressions. Suitable for students in Grades 9-10.

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