To solve for \((f/g)(x)\), we start by defining the functions \(f(x)\) and \(g(x)\):
\[ f(x) = 2x^2 - x - 6 \] \[ g(x) = x^2 + 3x - 10 \]
Now, to find \((f/g)(x)\), we compute:
\[ \frac{f(x)}{g(x)} = \frac{2x^2 - x - 6}{x^2 + 3x - 10} \]
Next, we will simplify this expression if possible, by factoring both the numerator and the denominator.
Step 1: Factor \(f(x)\)
We want to factor \(f(x) = 2x^2 - x - 6\). To factor, we look for two numbers that multiply to \(2 \times -6 = -12\) and add to \(-1\). The numbers \(-4\) and \(3\) work.
Now we rewrite:
\[ 2x^2 - 4x + 3x - 6 = 2x(x - 2) + 3(x - 2) = (2x + 3)(x - 2) \]
Thus,
\[ f(x) = (2x + 3)(x - 2) \]
Step 2: Factor \(g(x)\)
Now we factor \(g(x) = x^2 + 3x - 10\). We need numbers that multiply to \(-10\) and add to \(3\). The numbers \(5\) and \(-2\) work.
So we write:
\[ g(x) = (x + 5)(x - 2) \]
Step 3: Simplify \((f/g)(x)\)
Now we can rewrite \((f/g)(x)\):
\[ \frac{f(x)}{g(x)} = \frac{(2x + 3)(x - 2)}{(x + 5)(x - 2)} \]
We see that \((x - 2)\) cancels out.
Thus, we have:
\[ \frac{f(x)}{g(x)} = \frac{2x + 3}{x + 5} \quad (x \neq 2) \]
Step 4: Find the domain of \((f/g)(x)\)
The domain of \(\frac{f(x)}{g(x)}\) is restricted by the values that make \(g(x) = 0\). Setting \(g(x)\) to zero:
\[ g(x) = x^2 + 3x - 10 = 0 \]
We can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-10)}}{2 \cdot 1} = \frac{-3 \pm \sqrt{9 + 40}}{2} = \frac{-3 \pm \sqrt{49}}{2} = \frac{-3 \pm 7}{2} \]
Calculating the two possible values:
- \(x = \frac{4}{2} = 2\)
- \(x = \frac{-10}{2} = -5\)
Thus, the function \(g(x)\) is zero at \(x = 2\) and \(x = -5\). Therefore, \((f/g)(x)\) is undefined at these points.
Final Domain:
The domain of \((f/g)(x)\) is:
\[ \text{Domain} = { x \in \mathbb{R} , | , x \neq 2 , \text{and} , x \neq -5 } \]
So, in interval notation, the domain is:
\[ (-\infty, -5) \cup (-5, 2) \cup (2, \infty) \]