Answer
Quotient $=x^2-\dfrac{2x}{3}-\dfrac{1}{9}$
Remainder $=\dfrac{16x-17}{9}$
Work Step by Step
The given expression is
$(3x^4-x^3+x-2)\div(3x^2+x+1)$
Rewrite the expression as
$(3x^4-x^3+0x^2+x-2)\div(3x^2+x+1)$
Perform long division:
$\begin{matrix}
& x^2 & -2x/3 & -1/9 & & && \leftarrow &\text{Quotient}\\
&-- &-- &--&--& \\
3x^2+x+1) &3x^4&-x^3&+0x^2&+x&-2 & \\
& 3x^4 &+x^3 & +x^2& &&& \leftarrow &x^2(3x^2+x+1) \\
& -- & -- & --& &&&\leftarrow &\text{subtract} \\
& 0 & -2x^3& -x^2 &+x & & \\
& &-2x^3 & -2x^2/3&-2x/3 &&& \leftarrow &-2x/3(3x^2+x+1) \\
& & -- & --& -- &&&\leftarrow &\text{subtract} \\
& & 0 & -x^2/3& +5x/3 &-2 & & \\
& &&-x^2/3 & -x/9&-1/9 && \leftarrow &-1/9(3x^2+x+1) \\
& && -- & --& -- &&\leftarrow &\text{subtract} \\
& && 0 & +16x/9 & -17/9 & & \leftarrow & \text{Remainder}
\end{matrix}$
Checking:
$\text{(Quotient)(divisor)+ Remainder}$
$=(x^2-\frac{2x}{3}-\frac{1}{9})(3x^2+x+1)+\frac{16x}{9}-\frac{17}{9}$
$=3x^4-2x^3-\frac{x^2}{3}+x^3-\frac{2x^2}{3}-\frac{x}{9}+x^2-\frac{2x}{3}-\frac{1}{9}+\frac{16x}{9}-\frac{17}{9}$
$=3x^4-x^3+x-2$
$=\text{ Dividend}$
Hence, the quotient is $x^2-\frac{2x}{3}-\frac{1}{9}$ and the remainder is $\frac{16x}{9}-\frac{17}{9}$.
