Find the quotient. \[ \frac{2 x^{2}-12 x}{x^{2}-7 x+6} \div \frac{2 x}{3 x-3}=\square \]

To solve the expression \(\frac{2 x^{2}-12 x}{x^{2}-7 x+6} \div \frac{2 x}{3 x-3}\), we will first convert the division into multiplication by taking the reciprocal of the second fraction. 1. **Rewrite the expression**: \[ \frac{2 x^{2}-12 x}{x^{2}-7 x+6} \div \frac{2 x}{3 x-3} = \frac{2 x^{2}-12 x}{x^{2}-7 x+6} \cdot \frac{3 x-3}{2 x} \] 2. **Factor the polynomials**: - For the numerator of the first fraction \(2 x^{2}-12 x\): \[ 2 x(x - 6) \] - For the denominator of the first fraction \(x^{2}-7 x+6\): \[ (x - 1)(x - 6) \] - For the numerator of the second fraction \(3 x - 3\): \[ 3(x - 1) \] - The denominator of the second fraction \(2 x\) is already in simplest form. 3. **Substituting the factored forms**: \[ \frac{2 x(x - 6)}{(x - 1)(x - 6)} \cdot \frac{3(x - 1)}{2x} \] 4. **Canceling common factors**: - The \(2x\) in the numerator and denominator cancels. - The \(x - 6\) in the numerator and denominator cancels. - The \(x - 1\) in the numerator and denominator cancels. After canceling we have: \[ \frac{3}{1} = 3 \] Thus, the quotient is \(\boxed{3}\).