Question

Multiply the following and write your answer in lowest terms: $\frac{x^2-3x-10}{x^2-x-20}\times \frac{x^2-2x+4}{x^3+8}$

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

Answer 1

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

The given expression $\frac{x^2-3x-10}{x^2-x-20}\times \frac{x^2-2x+4}{x^3+8}$ can be simplified as shown below:

$=\frac{x^2-3x-10}{x^2-x-20}\times \frac{x^2-2x+4}{x^3+8}$

$=\frac{x^2-5x+2x-10}{x^2-5x+4x-20}\times \frac{x^2-2x+4}{x^3+2^3}$

$=\frac{x(x-5)+2(x-5)}{x(x-5)+4(x-5)}\times \frac{x^2-2x+4}{(x+2)(x^2+2^2-(x\times 2\left)\right)}(\because a^3+b^3=(a+b\left)\right(a^2+b^2-ab\left)\right)$

$=\frac{(x+2)(x-5)}{(x+4)(x-5)}\times \frac{x^2-2x+4}{(x+2)(x^2+4-2x)}$

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

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

$=\frac{1}{x+4}$

Hence, $\frac{x^2-3x-10}{x^2-x-20}\times \frac{x^2-2x+4}{x^3+8}=\frac{1}{x+4}$.