Answer
$3x^3 -7x^2+\frac{41x^2}{2x+6}$
Work Step by Step
$(6x^4+4x^3-x^2) \div (6+2x)$
$=(6x^4+4x^3-x^2) \div (2x+6)$
$3x^3 -7x^2$
_________________________
$2x+6)6x^4+4x^3-x^2$
$-6x^4-18x^3$
_________________________
$-14x^3-x^2$
$14x^3+42x^2$
_________________________
$41x^2$
The answer is $3x^3 -7x^2+\frac{41x^2}{2x+6}$