To find \((f/g)(x)\), we need to divide the function \(f(x)\) by the function \(g(x)\).
Given: \[ f(x) = 2x^2 - x - 6 \] \[ g(x) = x^2 + 3x - 10 \]
Now, we compute \((f/g)(x)\): \[ \frac{f(x)}{g(x)} = \frac{2x^2 - x - 6}{x^2 + 3x - 10} \]
Next, we can factor both \(f(x)\) and \(g(x)\) to see if there are any common factors that we can cancel.
Factoring \(f(x)\):
To factor \(f(x) = 2x^2 - x - 6\), we look for two numbers that multiply to \(2 \cdot (-6) = -12\) and add to \(-1\) (the coefficient of \(x\)).
The numbers \(-4\) and \(3\) work. We rewrite \(f(x)\): \[ f(x) = 2x^2 - 4x + 3x - 6 \] \[ = 2x(x - 2) + 3(x - 2) \] \[ = (2x + 3)(x - 2) \]
Factoring \(g(x)\):
To factor \(g(x) = x^2 + 3x - 10\), we need two numbers that multiply to \(-10\) and add to \(3\).
The numbers \(5\) and \(-2\) work. We rewrite \(g(x)\): \[ g(x) = (x + 5)(x - 2) \]
Now substituting back into \((f/g)(x)\):
Now we can reduce: \[ (f/g)(x) = \frac{(2x + 3)(x - 2)}{(x + 5)(x - 2)} \]
We see that \((x - 2)\) is a common factor in both the numerator and denominator: \[ (f/g)(x) = \frac{2x + 3}{x + 5} \quad \text{for } x \neq 2 \]
Finding the Domain:
The domain of the function \(\frac{f(x)}{g(x)}\) consists of all real numbers except where \(g(x) = 0\) since division by zero is undefined.
To find where \(g(x) = 0\): \[ g(x) = x^2 + 3x - 10 = 0 \] Factoring gives: \[ (x + 5)(x - 2) = 0 \] Setting each factor to zero gives us: \[ x + 5 = 0 \Rightarrow x = -5 \] \[ x - 2 = 0 \Rightarrow x = 2 \]
Therefore, the values where \(g(x)\) is zero are \(x = -5\) and \(x = 2\).
Final Domain:
Thus, the domain of the function \((f/g)(x)\) is all real numbers except \(x = -5\) and \(x = 2\). This can be written in interval notation as: \[ \text{Domain} = (-\infty, -5) \cup (-5, 2) \cup (2, \infty) \]