Answer
$$3x^3-4+\frac{1}{x}$$
Work Step by Step
$$\frac{9x^5\:-\:12x^2\:+\:3x}{3x^2}$$
Recall the process of dividing a polynomial by a monomial:
$$\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}, c\ne0$$
Thus, we have:
$$\frac{9x^5-12x^2+3x}{3x^2} = \frac{9x^5}{3x^2}-\frac{12x^2}{3x^2}+\frac{3x}{3x^2}$$ $$=3x^3-4+\frac{1}{x}$$
