Question

Divide the following and write your answer in lowest terms: $\frac{2x^2+5x-3}{2x^2+9x+9}÷\frac{2x^2+x-1}{2x^2+x-3}$

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

Answer 1

Answer: (A)

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

The given expression $\frac{2x^2+5x-3}{2x^2+9x+9}÷\frac{2x^2+x-1}{2x^2+x-3}$can be simplified as shown below:

$=\frac{2x^2+5x-3}{2x^2+9x+9}÷\frac{2x^2+x-1}{2x^2+x-3}$

$=\frac{2x^2+5x-3}{2x^2+9x+9}\times \frac{2x^2+x-3}{2x^2+x-1}$

$=\frac{2x^2+6x-x-3}{2x^2+6x+3x+9}\times \frac{2x^2-2x+3x-3}{2x^2+2x-x-1}$

$=\frac{2x(x+3)-1(x+3)}{2x(x+3)+3(x+3)}\times \frac{2x(x-1)+3(x-1)}{2x(x+1)-1(x+1)}$

$=\frac{(2x-1)(x+3)}{(2x+3)(x+3)}\times \frac{(2x+3)(x-1)}{(2x-1)(x+1)}$

$=\frac{(2x-1)(x+3)(2x+3)(x-1)}{(2x+3)(x+3)(2x-1)(x+1)}$

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

Hence, $\frac{2x^2+5x-3}{2x^2+9x+9}÷\frac{2x^2+x-1}{2x^2+x-3}=\frac{x-1}{x+1}$.