Answer
$3x^2+2x+3+\frac{(-6x+9)}{(x^2-1)}$
Work Step by Step
$3x^4+2x^3-8x+6/(x^2-1)$
$(x^2-1)*(3x^2)=3x^4-3x^2$
$3x^4+2x^3-8x+6-(3x^4-3x^2)$
$3x^4+2x^3-8x+6-3x^4+3x^2$
$2x^3+3x^2-8x+6$
$(x^2-1)*2x=2x^3-2x$
$2x^3+3x^2-8x+6-(2x^3-2x)$
$2x^3+3x^2-8x+6-2x^3+2x$
$3x^2-6x+6$
$(x^2-1)*3=3x^2-3$
$3x^2-6x+6-(3x^2-3)$
$3x^2-6x+6-3x^2+3$
$-6x+9$
$3x^2+2x+3+\frac{(-6x+9)}{(x^2-1)}$
