To find \( \frac{f(x)}{g(x)} \), we first need to define \( f(x) \) and \( g(x) \).
Given: \[ f(x) = 2x^2 - x - 6 \] \[ g(x) = x^2 + 3x - 10 \]
Now, we can calculate \( \frac{f(x)}{g(x)} \): \[ \frac{f(x)}{g(x)} = \frac{2x^2 - x - 6}{x^2 + 3x - 10} \]
Next, let's factor both the numerator and the denominator.
Factoring \( f(x) \):
To factor \( f(x) = 2x^2 - x - 6 \), we look for two numbers that multiply to \( 2 \times -6 = -12 \) and add to \( -1 \). These numbers are \( -4 \) and \( 3 \).
Thus, we can rewrite \( f(x) \): \[ f(x) = 2x^2 - 4x + 3x - 6 \] Group terms: \[ f(x) = (2x^2 - 4x) + (3x - 6) = 2x(x - 2) + 3(x - 2) \] Factoring by grouping gives: \[ f(x) = (2x + 3)(x - 2) \]
Factoring \( g(x) \):
Now, we factor \( g(x) = x^2 + 3x - 10 \). We look for two numbers that multiply to \( -10 \) and add to \( 3 \). These numbers are \( 5 \) and \( -2 \).
Thus, we can factor \( g(x) \): \[ g(x) = (x + 5)(x - 2) \]
Combining the results:
Now substituting the factored forms back into \( \frac{f(x)}{g(x)} \): \[ \frac{f(x)}{g(x)} = \frac{(2x + 3)(x - 2)}{(x + 5)(x - 2)} \]
Simplifying:
The term \( (x - 2) \) is common in both the numerator and denominator, so we cancel it out: \[ \frac{f(x)}{g(x)} = \frac{2x + 3}{x + 5} \] This is valid as long as \( x \neq 2 \).
Finding the domain:
The domain of \( \frac{f(x)}{g(x)} \) is all the values of \( x \) for which \( g(x) \neq 0 \).
Setting \( g(x) = x^2 + 3x - 10 = 0 \): Factoring gives us: \[ (x + 5)(x - 2) = 0 \] Thus, \[ x + 5 = 0 \quad \Rightarrow \quad x = -5 \] \[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \]
The function \( \frac{f(x)}{g(x)} \) is undefined at \( x = -5 \) and \( x = 2 \). Therefore, the domain of \( \frac{f(x)}{g(x)} \) is: \[ \text{Domain: } x \in \mathbb{R} \setminus {-5, 2} \]
So, in interval notation, the domain of \( \frac{f(x)}{g(x)} \) can be expressed as: \[ (-\infty, -5) \cup (-5, 2) \cup (2, \infty) \]