Question

Multiply the following and write your answer in lowest terms: $\frac{x^2-81}{x^2-4}\times \frac{x^2+6x+8}{x^2-5x-36}$

• A. $\frac{x+9}{x-2}$
• B. $\frac{x+4}{x-2}$
• C. $\frac{x+9}{x-9}$
• D. None of these

Answer 1

Answer: (A)

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

The given expression $\frac{x^2-81}{x^2-4}\times \frac{x^2+6x+8}{x^2-5x-36}$ can be simplified as shown below:

$=\frac{x^2-81}{x^2-4}\times \frac{x^2+6x+8}{x^2-5x-36}$

$=\frac{x^2-9^2}{x^2-2^2}\times \frac{x^2+4x+2x+8}{x^2-9x+4x-36}$

$=\frac{(x-9)(x+9)}{(x-2)(x+2)}\times \frac{x(x+4)+2(x+4)}{x(x-9)+4(x-9)}(\because (a+b{)}^2=a^2+b^2+2ab)$

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

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

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

Hence, $\frac{x^2-81}{x^2-4}\times \frac{x^2+6x+8}{x^2-5x-36}=\frac{x+9}{x-2}$.