Find the quotient. \( \frac{8 x}{3 x+18} \div \frac{x^{2}-6 x}{x^{2}-36}=\square, x \neq \square, x \neq \square, x \neq \square \)

To solve the expression \( \frac{8 x}{3 x + 18} \div \frac{x^{2}-6 x}{x^{2}-36} \), we will first transform the division into multiplication by taking the reciprocal of the second fraction. 1. Rewrite the problem: \[ \frac{8 x}{3 x + 18} \cdot \frac{x^{2}-36}{x^{2}-6 x} \] 2. Factor the terms: - The denominator \(3x + 18\) can be factored out as \(3(x + 6)\). - The numerator \(x^2 - 36\) is a difference of squares, so it factors to \((x - 6)(x + 6)\). - The numerator \(x^2 - 6x\) can be factored as \(x(x - 6)\). Now, we have: \[ \frac{8 x}{3(x + 6)} \cdot \frac{(x - 6)(x + 6)}{x(x - 6)} \] 3. Cancel out common terms: - The \(x\) in the numerator and denominator cancels. - The \((x - 6)\) in the numerator and denominator cancels. Now it simplifies to: \[ \frac{8(x + 6)}{3(x + 6)} \] Assuming \(x + 6 \neq 0\) (which excludes \(x = -6\)), we can cancel \((x + 6)\): \[ \frac{8}{3} \] Thus, the final simplified answer is: \[ \frac{8}{3}, \quad x \neq -6, \quad x \neq 0, \quad x \neq 6 \]