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

1. Write the division as a multiplication by the reciprocal: \[ \frac{8x}{3x+18} \div \frac{x^2-6x}{x^2-36} = \frac{8x}{3x+18} \times \frac{x^2-36}{x^2-6x} \] 2. Factor each polynomial where possible: - For \(3x+18\): \[ 3x+18 = 3(x+6) \] - For \(x^2-36\) (a difference of squares): \[ x^2-36 = (x-6)(x+6) \] - For \(x^2-6x\): \[ x^2-6x = x(x-6) \] 3. Substitute the factorizations into the expression: \[ \frac{8x}{3(x+6)} \times \frac{(x-6)(x+6)}{x(x-6)} \] 4. Cancel common factors (note the restrictions for cancellation): - The factor \(x\) cancels with \(x\) (provided \(x\neq 0\)). - The factor \((x+6)\) cancels with \((x+6)\) (provided \(x\neq -6\)). - The factor \((x-6)\) cancels with \((x-6)\) (provided \(x\neq 6\)). After canceling, the expression simplifies to: \[ \frac{8}{3} \] 5. Determine the restrictions from the original expression: - \(3x+18 \neq 0 \implies x \neq -6\) - \(x^2-36 \neq 0 \implies x \neq 6 \; \text{and} \; x \neq -6\) - \(x^2-6x \neq 0\) (since it is the divisor after inversion) \(\implies x(x-6) \neq 0 \implies x \neq 0 \; \text{and} \; x \neq 6\) Combining these restrictions: \[ x \neq -6,\quad x \neq 0,\quad x \neq 6 \] 6. Final answer: \[ \frac{8x}{3x+18} \div \frac{x^2-6x}{x^2-36} = \frac{8}{3}, \quad x \neq -6,\ x \neq 0,\ x \neq 6. \]