Question

Divide the following and write your answer in lowest terms: $\frac{2x^2+13x+15}{x^2+3x-10}÷\frac{2x^2-x-6}{x^2-4x+4}$

Answer 1

The given expression $\frac{2x^2+13x+15}{x^2+3x-10}÷\frac{2x^2-x-6}{x^2-4x+4}$can be simplified as shown below:

$=\frac{2x^2+13x+15}{x^2+3x-10}÷\frac{2x^2-x-6}{x^2-4x+4}$

$=\frac{2x^2+13x+15}{x^2+3x-10}\times \frac{x^2-4x+4}{2x^2-x-6}$

$=\frac{2x^2+10x+3x+15}{x^2+5x-2x-10}\times \frac{x^2-2x-2x+4}{2x^2-4x+3x-6}$

$=\frac{2x(x+5)+3(x+5)}{x(x+5)-2(x+5)}\times \frac{x(x-2)-2(x-2)}{2x(x-2)+3(x-2)}$

$=\frac{(2x+3)(x+5)}{(x-2)(x+5)}\times \frac{(x-2)(x-2)}{(2x+3)(x-2)}$

$=\frac{(2x+3)(x+5)(x-2)(x-2)}{(x-2)(x+5)(2x+3)(x-2)}$

$=1$

Hence, $\frac{2x^2+13x+15}{x^2+3x-10}÷\frac{2x^2-x-6}{x^2-4x+4}=1$.