## Intermediate Algebra (6th Edition)

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