Question

Divide the following and write your answer in lowest terms: $\frac{x^2-4x-5}{x^2-25}÷\frac{x^2-3x-10}{x^2+7x+10}$

• A. $\frac{x+1}{x-5}$
• B. $\frac{x-1}{x-5}$
• C. $\frac{x+1}{x+5}$
• D. None of these

Answer 1

Answer: (A)

$\frac{x+1}{x-5}$

The given expression $\frac{x^2-4x-5}{x^2-25}÷\frac{x^2-3x-10}{x^2+7x+10}$can be simplified as shown below:

$=\frac{x^2-4x-5}{x^2-25}÷\frac{x^2-3x-10}{x^2+7x+10}$

$=\frac{x^2-4x-5}{x^2-25}\times \frac{x^2+7x+10}{x^2-3x-10}$

$=\frac{x^2-5x+x-5}{x^2-5^2}\times \frac{x^2+5x+2x+10}{x^2-5x+2x-10}$

$=\frac{x(x-5)+1(x-5)}{(x-5)(x+5)}\times \frac{x(x+5)+2(x+5)}{x(x-5)+2(x-5)}(\because a^2-b^2=(a-b\left)\right(a+b\left)\right)$

$=\frac{(x+1)(x-5)}{(x-5)(x+5)}\times \frac{(x+2)(x+5)}{(x+2)(x-5)}$

$=\frac{(x+1)(x-5)(x+2)(x+5)}{(x-5)(x+5)(x+2)(x-5)}$

$=\frac{x+1}{x-5}$

Hence, $\frac{x^2-4x-5}{x^2-25}÷\frac{x^2-3x-10}{x^2+7x+10}=\frac{x+1}{x-5}$.