Question

f(x)=2x^2-3x-9
g(x)=4x^2-9
find (f/g)(x)
a)x-3/2x+3
b)1/2+1/3x+1
c)x+3/2x-3
d)x-3/2x-3

Answer 1

To find \((f/g)(x)\), we need to compute the quotient of the two functions \(f(x)\) and \(g(x)\):

1. **Define the functions**:
\[
f(x) = 2x^2 - 3x - 9
\]
\[
g(x) = 4x^2 - 9
\]

2. **Calculate \((f/g)(x)\)**:
\[
(f/g)(x) = \frac{f(x)}{g(x)} = \frac{2x^2 - 3x - 9}{4x^2 - 9}
\]

3. **Factor the numerator and the denominator**:
- **Factor \(g(x)\)**:
\[
g(x) = 4x^2 - 9 = (2x - 3)(2x + 3) \quad \text{(difference of squares)}
\]

- **Factor \(f(x)\)**: We will factor \(2x^2 - 3x - 9\).
To factor it, we need to find two numbers that multiply to \(2 \cdot (-9) = -18\) and add to \(-3\).
These numbers are \(-6\) and \(3\).
\[
f(x) = 2x^2 - 6x + 3x - 9 = 2x(x - 3) + 3(x - 3) = (2x + 3)(x - 3)
\]

4. **Now we can rewrite \((f/g)(x)\)**:
\[
(f/g)(x) = \frac{(2x + 3)(x - 3)}{(2x - 3)(2x + 3)}
\]
The \((2x + 3)\) in the numerator and denominator cancels out, resulting in:
\[
(f/g)(x) = \frac{x - 3}{2x - 3} \quad \text{(valid as long as \(x \neq -\frac{3}{2}\))}
\]

5. **Final Result**:
The simplified form is:
\[
(f/g)(x) = \frac{x - 3}{2x - 3}
\]

Thus, the correct answer is:
**d) \(\frac{x - 3}{2x - 3}\)**.