Answer:
[tex]3x^2-6-\dfrac{2x-13}{x^2+2}[/tex]
Step-by-step explanation:
As the divisor is quadratic, use long division rather than synthetic division:
[tex]\large \begin{array}{r}3x^2-6\phantom{)}\\x^2+2{\overline{\smash{\big)}\,3x^4-2x+1\phantom{)}}}\\{-~\phantom{(}\underline{(3x^4+6x^2)\phantom{-b)}}\\-6x^2-2x+1\phantom{)}\\-~\phantom{()}\underline{(-6x^2\phantom{))))}-12)\phantom{}}\\-2x+13\phantom{)}\\\end{array}[/tex]
Therefore:
[tex]\textsf{Dividend}: \quad 3x^4-2x+1[/tex]
[tex]\textsf{Divisor}: \quad x^2+2[/tex]
[tex]\textsf{Quotient}: \quad 3x^2-6[/tex]
[tex]\textsf{Remainder}: \quad -2x+13=-(2x-13)[/tex]
When dividing a polynomial, the result is the quotient plus the remainder over the divisor:
[tex]\implies \dfrac{3x^4-2x+1}{x^2+2}=3x^2-6-\dfrac{2x-13}{x^2+2}[/tex]
