Question

multiply the following rational expressions.
\\(\frac{x^2 - 16}{x^2 - 8x + 16} \cdot \frac{2x - 8}{x + 4}\\)

select the correct choice below and fill in the answer box to complete your choice.
(simplify your answer.)

\\(\bigcirc\\) a. \\(\frac{x^2 - 16}{x^2 - 8x + 16} \cdot \frac{2x - 8}{x + 4} = \square, x \
eq -4, 4\\)

\\(\bigcirc\\) b. \\(\frac{x^2 - 16}{x^2 - 8x + 16} \cdot \frac{2x - 8}{x + 4} = \square, x \
eq -16, -4, 8\\)

Explanation:

Step1: Factor all expressions

$x^2-16=(x-4)(x+4)$
$x^2-8x+16=(x-4)^2$
$2x-8=2(x-4)$

Step2: Rewrite product with factors

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

Step3: Cancel common factors

Cancel $(x-4)$, $(x+4)$, and another $(x-4)$ from numerator/denominator:
$\frac{\cancel{(x-4)}\cancel{(x+4)}}{\cancel{(x-4)}^2} \cdot \frac{2\cancel{(x-4)}}{\cancel{x+4}} = 2$

Step4: Identify restricted values

Original denominators: $x^2-8x+16=(x-4)^2$ cannot be 0, so $x
eq 4$.
$x+4$ cannot be 0, so $x
eq -4$.
Canceled factors also restrict $x
eq 4, -4$.

Answer:

A. $\frac{x^2 - 16}{x^2 - 8x + 16} \cdot \frac{2x - 8}{x + 4} = 2$, $x
eq - 4,4$