Appendix C - Synthetic Division and the Remainder Theorem - C Exercise Set - Page 992: 15

$ 3y^2+2y+6+\dfrac{-2}{y-3}$

Equating to zero the divisor of the given expression, $( 3y^3-7y^2-20 )\div( y-3 ) ,$ and then solving for the variable, result to \begin{align*} y-3&=0 \\ y&=3 .\end{align*} Using $ 3 $ in operating the coefficients of the dividend through synthetic division gives the result shown below. The last row of numbers, $\left\{ 3,2,6,-2 \right\},$ gives the coefficient of the quotient and the remainder. Hence, $( 3y^3-7y^2-20 )\div( y-3 ) $ is equal to \begin{align*} 3y^2+2y+6+\dfrac{-2}{y-3} .\end{align*}