What is the product? \( \frac{2 a-7}{a} \cdot \frac{3 a^{2}}{2 a^{2}-11 a+14} \)

To find the product of the two fractions, we begin by multiplying the numerators together and the denominators together. The expression can be simplified by factoring. The first fraction is \( \frac{2a - 7}{a} \) and the second fraction is \( \frac{3a^2}{2a^2 - 11a + 14} \). First, we factor the denominator of the second fraction. We need to factor \( 2a^2 - 11a + 14 \). We can look for two numbers that multiply to \( 2 \times 14 = 28 \) and add up to \( -11 \). These numbers are \( -7 \) and \( -4 \). So we can write: \[ 2a^2 - 11a + 14 = 2a^2 - 7a - 4a + 14 = (2a - 7)(a - 2) \] Now we have: \[ \frac{2a - 7}{a} \cdot \frac{3a^2}{(2a - 7)(a - 2)} \] Next, we can cancel out the \( 2a - 7 \) in the numerator and denominator: \[ = \frac{3a^2}{a(a - 2)} \] Now we can further simplify \( \frac{3a^2}{a(a - 2)} \): \[ = \frac{3a}{a - 2} \quad \text{(as long as } a \neq 0\text{)} \] So the final answer is: \[ \frac{3a}{a - 2} \quad \text{(as long as } a \neq 0\text{ and } a \neq 2\text{)} \]