## Intermediate Algebra (6th Edition)

$\dfrac{x-4}{2}$
Factoring the given expressions and cancelling the common factors between the numerator and the denominator, the given rational expression, $\dfrac{x^2-8x+16}{2x-8} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{(x-4)(x-4)}{2(x-4)} \\\\= \dfrac{(\cancel{x-4})(x-4)}{2(\cancel{x-4})} \\\\= \dfrac{x-4}{2} .\end{array}