Answer
$\dfrac{6x+1}{3}$
Work Step by Step
Factoring the expressions and cancelling the common factors between the numerator and the denominator, the given expression, $
(6x^2-17x-3)\div(3x-9)
,$ simplifies to
\begin{array}{l}\require{cancel}
\dfrac{6x^2-17x-3}{3x-9}
\\\\=
\dfrac{(6x+1)(x-3)}{3(x-3)}
\\\\=
\dfrac{(6x+1)(\cancel{x-3})}{3(\cancel{x-3})}
\\\\=
\dfrac{6x+1}{3}
.\end{array}
