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

We start with the expression \[ \frac{8x}{3x+18} \div \frac{x^2-6x}{x^2-36}. \] **Step 1. Rewrite the division as multiplication by the reciprocal.** \[ \frac{8x}{3x+18} \div \frac{x^2-6x}{x^2-36} = \frac{8x}{3x+18} \cdot \frac{x^2-36}{x^2-6x}. \] **Step 2. Factor all expressions where possible.** - Factor the denominator \(3x+18\): \[ 3x+18 = 3(x+6). \] - Factor \(x^2-6x\) in the denominator of the second fraction: \[ x^2-6x = x(x-6). \] - Factor \(x^2-36\) in the numerator of the second fraction (difference of two squares): \[ x^2-36 = (x-6)(x+6). \] Substitute these factorizations back into the expression: \[ \frac{8x}{3(x+6)} \cdot \frac{(x-6)(x+6)}{x(x-6)}. \] **Step 3. Cancel common factors.** We see that common factors can be canceled provided they are not zero: - The factor \(x\) cancels (provided \(x \neq 0\)). - The factor \(x+6\) cancels (provided \(x \neq -6\)). - The factor \(x-6\) cancels (provided \(x \neq 6\)). After canceling, we are left with: \[ \frac{8}{3}. \] **Step 4. Determine the restrictions on \(x\).** From the original expression, we must have: 1. \(3x+18 \neq 0 \Rightarrow x+6 \neq 0 \Rightarrow x \neq -6\). 2. \(x^2-36 \neq 0 \Rightarrow x \neq 6\) and \(x \neq -6\). 3. The divisor \(\frac{x^2-6x}{x^2-36}\) must not be zero. Its numerator \(x^2-6x = x(x-6)\) must not equal zero, so \[ x \neq 0 \quad \text{and} \quad x \neq 6. \] Thus, the restrictions are: \[ x \neq -6, \quad x \neq 0, \quad x \neq 6. \] **Final Answer:** \[ \frac{8x}{3x+18} \div \frac{x^2-6x}{x^2-36} = \frac{8}{3}, \quad x \neq -6,\; x \neq 0,\; x \neq 6. \]