Answer
$-\dfrac{(x-4)^2}{4x}, x\ne-4, 0, 4$
Work Step by Step
The initila restrictions are $x\ne-4, 4$.
Use the rule $\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c}$ to obtain:
$\dfrac{4-x}{4+x} \cdot \dfrac{x^2-16}{4x}$
$=\dfrac{4-x}{4+x} \cdot \dfrac{x^2-4^2}{4x}$
$=\dfrac{-1(-4+x)}{4+x} \cdot \dfrac{x^2-4^2}{4x}$
$=\dfrac{-(x-4)}{4+x} \cdot \dfrac{x^2-4^2}{4x}$
There is an additional restriction $x\ne0$.
Factor each polynomial completely.
Use special formula $a^2-b^2=(a+b)(a-b)$ to obtain:
$=\dfrac{-(x-4)}{4+x} \cdot \dfrac{(x+4)(x-4)}{4x}, x\ne-4, 0, 4$
Cancel the common factor $x-4$:
$=-\dfrac{(x-4)(x-4)}{4x}$
$=-\dfrac{(x-4)^2}{4x}, x\ne-4, 0, 4$
Hence, the lowest term is:
$-\dfrac{(x-4)^2}{4x}, x\ne-4, 0, 4$
