Question

Divide the following and write your answer in lowest terms: $\frac{x^2+11x+28}{x^2-4x-77}÷\frac{x^2+7x+12}{x^2-2x-15}$

• A. $\frac{x-5}{x-11}$
• B. $\frac{x+5}{x-11}$
• C. $\frac{x-5}{x+11}$
• D. None of these

Answer 1

Answer: (A)

$\frac{x-5}{x-11}$

The given expression $\frac{x^2+11x+28}{x^2-4x-77}÷\frac{x^2+7x+12}{x^2-2x-15}$can be simplified as shown below:

$=\frac{x^2+11x+28}{x^2-4x-77}÷\frac{x^2+7x+12}{x^2-2x-15}$

$=\frac{x^2+11x+28}{x^2-4x-77}\times \frac{x^2-2x-15}{x^2+7x+12}$

$=\frac{x^2+7x+4x+28}{x^2-11x+7x-77}\times \frac{x^2-5x+3x-15}{x^2+4x+3x+12}$

$=\frac{x(x+7)+4(x+7)}{x(x-11)+7(x-11)}\times \frac{x(x-5)+3(x-5)}{x(x+4)+3(x+4)}$

$=\frac{(x+4)(x+7)}{(x+7)(x-11)}\times \frac{(x+3)(x-5)}{(x+3)(x+4)}$

$=\frac{(x+4)(x+7)(x+3)(x-5)}{(x+7)(x-11)(x+3)(x+4)}$

$=\frac{x-5}{x-11}$

Hence, $\frac{x^2+11x+28}{x^2-4x-77}÷\frac{x^2+7x+12}{x^2-2x-15}=\frac{x-5}{x-11}$.