To find \( \frac{f}{g}(x) \), we first need to express \( f(x) \) and \( g(x) \):
- \( f(x) = 2x^2 - x - 6 \)
- \( g(x) = x^2 + 3x - 10 \)
Step 1: Calculate \( \frac{f}{g}(x) \)
\[ \frac{f}{g}(x) = \frac{f(x)}{g(x)} = \frac{2x^2 - x - 6}{x^2 + 3x - 10} \]
Step 2: Factor \( f(x) \) and \( g(x) \)
Factoring \( f(x) \):
To factor \( 2x^2 - x - 6 \), we need two numbers that multiply to \( 2 \cdot (-6) = -12 \) and add to \(-1\). The numbers \(-4\) and \(3\) work.
\[ f(x) = 2x^2 - 4x + 3x - 6 = 2x(x - 2) + 3(x - 2) = (2x + 3)(x - 2) \]
Factoring \( g(x) \):
To factor \( x^2 + 3x - 10 \), we need two numbers that multiply to \(-10\) and add to \(3\). The numbers \(5\) and \(-2\) work.
\[ g(x) = x^2 + 5x - 2x - 10 = (x + 5)(x - 2) \]
Step 3: Rewrite \( \frac{f}{g}(x) \)
Using the factored forms:
\[ \frac{f}{g}(x) = \frac{(2x + 3)(x - 2)}{(x + 5)(x - 2)} \]
We can simplify this fraction:
\[ \frac{f}{g}(x) = \frac{2x + 3}{x + 5}, \quad \text{for } x \neq 2 \]
Step 4: Determine the Domain of \( \frac{f}{g}(x) \)
The domain of \( \frac{f}{g}(x) \) is all \( x \) values for which \( g(x) \) is not zero.
To find where \( g(x) = 0 \):
Set \( g(x) = x^2 + 3x - 10 = 0 \). Factoring gives:
\[ (x + 5)(x - 2) = 0 \]
This gives the roots:
- \( x + 5 = 0 \) ⇒ \( x = -5 \)
- \( x - 2 = 0 \) ⇒ \( x = 2 \)
Thus, \( g(x) \) is zero at \( x = -5 \) and \( x = 2 \).
Therefore, the domain of \( \frac{f}{g}(x) \) is:
\[ \text{Domain: } x \in \mathbb{R} \text{, where } x \neq -5 \text{ and } x \neq 2. \]
In interval notation, the domain is:
\[ (-\infty, -5) \cup (-5, 2) \cup (2, \infty) \]