Question

simplify.
\frac{x^2 - 3x - 18}{-2x} \cdot \frac{x^2 + 9x}{x^2 - 9} \div \frac{x - 6}{x - 3}

Explanation:

Step1: Rewrite division as multiplication

$\frac{x^2 - 3x - 18}{2x} \cdot \frac{x^2 + 9x}{x^2 - 9} \cdot \frac{x - 3}{x - 6}$

Step2: Factor all polynomials

$\frac{(x-6)(x+3)}{2x} \cdot \frac{x(x+9)}{(x-3)(x+3)} \cdot \frac{x - 3}{x - 6}$

Step3: Cancel common factors

Cancel $(x-6)$, $(x+3)$, $x$, $(x-3)$ from numerator and denominator
$\frac{\cancel{(x-6)}\cancel{(x+3)}}{2\cancel{x}} \cdot \frac{\cancel{x}(x+9)}{\cancel{(x-3)}\cancel{(x+3)}} \cdot \frac{\cancel{x - 3}}{\cancel{x - 6}} = \frac{x+9}{2}$

Answer:

$\frac{x+9}{2}$