## Intermediate Algebra (6th Edition)

$3x+4$
Factoring the expressions and cancelling the common factors between the numerator and the denominator, the given expression, $(3x^2+19x+20)\div(x+5) ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{3x^2+19x+20}{x+5} \\\\= \dfrac{(3x+4)(x+5)}{x+5} \\\\= \dfrac{(3x+4)(\cancel{x+5})}{\cancel{x+5}} \\\\= 3x+4 .\end{array}