Answer
$\frac{x^4(x+1)}{3(x^2+1)}$
Work Step by Step
Step 1. Change the division to multiplication and factor the expressions, we have $\frac{5x^2-9x-2}{30x^3+6x^2}\div\frac{x^4-3x^2-4}{2x^8+6x^7+4x^6}=\frac{5x^2-9x-2}{30x^3+6x^2}\times\frac{2x^8+6x^7+4x^6}{x^4-3x^2-4}=\frac{(5x+1)(x-2)}{6x^2(5x+1)}\times\frac{2x^6(x^2+3x+2)}{(x^2-4)(x^2+1)}$
Step 2. Cancel common factors and continue factoring, we have $\frac{(5x+1)(x-2)}{6x^2(5x+1)}\times\frac{2x^6(x^2+3x+2)}{(x^2-4)(x^2+1)}=\frac{(x-2)}{3}\times\frac{x^4(x+1)(x+2)}{(x+2)(x-2)(x^2+1)}
=\frac{1}{3}\times\frac{x^4(x+1)}{(x^2+1)}=\frac{x^4(x+1)}{3(x^2+1)}$
