Intermediate Algebra (6th Edition)

$\dfrac{2x-1}{2}$
Factoring the expressions and cancelling the common factors between the numerator and the denominator, the given expression, $(2x^2+3x-2)\div(2x+4) ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{2x^2+3x-2}{2x+4} \\\\= \dfrac{(2x-1)(x+2)}{2(x+2)} \\\\= \dfrac{(2x-1)(\cancel{x+2})}{2(\cancel{x+2})} \\\\= \dfrac{2x-1}{2} .\end{array}