Question

perform the following operation and express in simplest form.\\(\dfrac{x^2 - 16}{2x - 18} \div \dfrac{x^2 - 2x - 8}{x + 2}\\)

Explanation:

Step1: Rewrite division as multiplication

$\frac{x^2 - 16}{2x - 18} \times \frac{x + 2}{x^2 - 2x - 8}$

Step2: Factor all polynomials

$\frac{(x-4)(x+4)}{2(x-9)} \times \frac{x+2}{(x-4)(x+2)}$

Step3: Cancel common factors

$\frac{\cancel{(x-4)}(x+4)}{2(x-9)} \times \frac{\cancel{x+2}}{\cancel{(x-4)}\cancel{(x+2)}}$

Step4: Multiply remaining terms

$\frac{x+4}{2(x-9)}$

Answer:

$\frac{x+4}{2(x-9)}$