Appendix C - Synthetic Division and the Remainder Theorem - C Exercise Set - Page 993: 19

$3x^2+6x-3+\dfrac{6}{x+\frac{1}{3}}$

Equating the divisor to zero of the given expression, $( 3x^3+1-x+7x^2 )\div( x+\dfrac{1}{3} )$ which is equivalent to $( 3x^3+7x^2-x+1 )\div( x+\dfrac{1}{3} ),$ and then solving for the variable, then \begin{align*} x+\dfrac{1}{3}&=0 \\ x&=-\dfrac{1}{3} .\end{align*} Using $ -\dfrac{1}{3} $ in manipulating the coefficients of the dividend results to the picture shown below. The last row of numbers, $\{ 3,6,-3,2 \},$ represent the coefficients of the quotient and the remainder. However, there is a need to multiply the remainder by $3$ since the number used to manipulate the coefficients, $-\dfrac{1}{3},$ was divided by $3$. Hence, $( 3x^3+1-x+7x^2 )\div( x+\dfrac{1}{3} )$ is equal to \begin{align*}&3x^2+6x-3+\dfrac{2(3)}{x+\frac{1}{3}} \\\\&= 3x^2+6x-3+\dfrac{6}{x+\frac{1}{3}} .\end{align*}