Answer
$\dfrac{x^4(x+1)}{3(x^2+1)}$
Work Step by Step
Factoring the expressions and then cancelling the common factors between the numerator and the denominator, the given expression, $
\dfrac{5x^2-9x-2}{30x^3+6x^2}\div\dfrac{x^4-3x^2-4}{2x^8+6x^7+4x^6}
,$ simplifies to
\begin{array}{l}\require{cancel}
\dfrac{5x^2-9x-2}{30x^3+6x^2}\cdot\dfrac{2x^8+6x^7+4x^6}{x^4-3x^2-4}
\\\\=
\dfrac{(5x+1)(x-2)}{6x^2(5x+1)}\cdot\dfrac{2x^6(x^2+3x+2)}{(x^2-4)(x^2+1)}
\\\\=
\dfrac{(5x+1)(x-2)}{6x^2(5x+1)}\cdot\dfrac{2x^6(x+2)(x+1)}{(x+2)(x-2)(x^2+1)}
\\\\=
\dfrac{(\cancel{5x+1})(\cancel{x-2})}{\cancel{2}(3)\cancel{x^2}(\cancel{5x+1})}\cdot\dfrac{\cancel{2}\cancel{x^2}(x^4)(\cancel{x+2})(x+1)}{(\cancel{x+2})(\cancel{x-2})(x^2+1)}
\\\\=
\dfrac{x^4(x+1)}{3(x^2+1)}
.\end{array}
