Question

Divide the following and write your answer in lowest terms: $\frac{3x^2-x-4}{9x^2-16}÷\frac{4x^2-4}{3x^2-2x-1}$

• A. $\frac{3x+1}{4(3x+4)}$
• B. $\frac{3x-1}{4(3x+4)}$
• C. $\frac{3x+1}{4(3x-4)}$
• D. None of these

Answer 1

The correct option is A $\frac{3x+1}{4(3x+4)}$

The given expression $\frac{3x^2-x-4}{9x^2-16}÷\frac{4x^2-4}{3x^2-2x-1}$can be simplified as shown below:

$=\frac{3x^2-x-4}{9x^2-16}÷\frac{4x^2-4}{3x^2-2x-1}$

$=\frac{3x^2-x-4}{9x^2-16}\times \frac{3x^2-2x-1}{4x^2-4}$

$=\frac{3x^2+3x-4x-4}{(3x{)}^2-4^2}\times \frac{3x^2-3x+x-1}{4(x^2-1)}$

$=\frac{3x(x+1)-4(x+1)}{(3x-4)(3x+4)}\times \frac{3x(x-1)+1(x-1)}{4(x-1)(x+1)}(\because a^2-b^2=(a-b\left)\right(a+b\left)\right)$

$=\frac{(3x-4)(x+1)}{(3x-4)(3x+4)}\times \frac{(3x+1)(x-1)}{4(x-1)(x+1)}$

$=\frac{(3x-4)(x+1)(3x+1)(x-1)}{4(3x-4)(3x+4)(x-1)(x+1)}$

$=\frac{3x+1}{4(3x+4)}$

Hence, $\frac{3x^2-x-4}{9x^2-16}÷\frac{4x^2-4}{3x^2-2x-1}=\frac{3x+1}{4(3x+4)}$.