Question

Express the following as a rational expression in lowest terms: $\frac{x^3-8}{x^2+4x-12}÷\frac{x^2+2x+4}{x+6}$

• A. $\frac{(x+4)}{(x-3)}$
• B. $\frac{x}{x-4}$
• C. $\frac{x-4}{x-3}$
• D. 1

Answer 1

Answer: (D)

$\frac{x^3-8}{x^2+4x-12}÷\frac{x^2+2x+4}{x+6}$
$\frac{\left(x^3-8\right)}{x^2+6x-2x-12}÷\frac{x^2+2x+4}{x+6}$
$=\frac{(x-2)\left(x^2+2x+4\right)}{(x+6)(x-2)}\times \frac{(x+6)}{\left(x^2+2x+4\right)}=1$