Question

Multiply the following and write your answer in lowest terms:

$\frac{2x-1}{x^2+2x+4}\times \frac{x^4-8x}{2x^2+5x-3}\times \frac{x+3}{x^2-2x}$

Answer 1

The given expression $\frac{2x-1}{x^2+2x+4}\times \frac{x^4-8x}{2x^2+5x-3}\times \frac{x+3}{x^2-2x}$can be simplified as shown below:

$=\frac{2x-1}{x^2+2x+4}\times \frac{x^4-8x}{2x^2+5x-3}\times \frac{x+3}{x^2-2x}$

$=\frac{2x-1}{x^2+2x+4}\times \frac{x(x^3-8)}{2x^2+6x-x-3}\times \frac{x+3}{x(x-2)}$

$=\frac{2x-1}{x^2+2x+4}\times \frac{x(x^3-2^3)}{2x^2+6x-x-3}\times \frac{x+3}{x(x-2)}$

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

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

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

$=1$

Hence, $\frac{2x-1}{x^2+2x+4}\times \frac{x^4-8x}{2x^2+5x-3}\times \frac{x+3}{x^2-2x}=1$.