Question

Solve $\left(\frac{x^2-2x-8}{x^2-4}\right)\left(\frac{6-3x}{20-5x}\right)=$

• A. $\frac{-3}{5}$
• B. $\frac{3}{5}$
• C. $\frac{3(x+4)(2-x)}{5(x+2)(4-x)}$
• D. $\frac{3(x+4)(2-x)}{5(x+2)(x-4)}$
• E. $\frac{3(x+4)}{5(4-x)}$

Answer 1

Answer: (B)

$\frac{3}{5}$

The value of $\left(\frac{x^2-2x-8}{x^2-4}\right)\left(\frac{6-3x}{20-5x}\right)$

$\Rightarrow \left(\frac{x^2-4x+2x-8}{(x+2)(x+2)}\right)\left(\frac{3(2-x)}{4(5-x)}\right)$

$\Rightarrow \left(\frac{x(x-4)+2(x-4)}{(x+2)(x-2)}\right)\left(\frac{3(2-x)}{5(4-x)}\right)$

$\Rightarrow \left(\frac{(x-4)(x+2)}{(x+2)(x-2)}\right)\left(\frac{3(x-2)}{5(x-4)}\right)$

$\Rightarrow \left(\frac{x-4}{x-2}\right)\left(\frac{3(x-2)}{5(x-4)}\right)=\frac{3}{5}$