Answer
$\dfrac{3}{x-2}, x\ne -2, 0, 2$
Work Step by Step
The initial restrictions is $x\ne0$.
Use the rule $\frac{a}{b}\div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c}$ to obtain:
$\dfrac{x-2}{4x} \cdot \dfrac{12x}{x^2-4x+4}$
$=\dfrac{x-2}{4x} \cdot \dfrac{12x}{x^2-2(x)(2)+2^2}$
Factor each polynomial completely.
Use special formula $a^2-2ab+b^2=(a-b)^2$ to obtain:
$=\dfrac{x-2}{4x} \cdot \dfrac{3\cdot 4x}{(x-2)^2}$
$=\dfrac{x-2}{4x} \cdot \dfrac{3\cdot 4x}{(x-2)(x-2)}$
This yields an additional restriction $x\ne \pm2$.
Cancel the common factor $x-2$ to obtain:.
$=\dfrac{3}{x-2}, x\ne -2, 0, 2$
Hence, the lowest term is:
$\dfrac{3}{x-2}, \space x\ne -2, 0, 2$
