When $3y^3-13y^2+11y+23$ is divided by $3y+2$, the quotient is $y^2-5y+c$ and the remainder is constant. Find the remainder.
To find the remainder, I need to use the remainder theorem, which says that the remainder of a polynomial $p(y)$ divided by $y-a$ is equal to $p(a)$. In this case, the divisor is $3y+2$, so I need to find a value of $y$ that makes it zero. I can do this by setting $3y+2=0$ and solving for $y$. I get $y=-\frac{2}{3}$. This means that the remainder is equal to $p(-\frac{2}{3})$, where $p(y)=3y^3-13y^2+11y+23$. To evaluate $p(-\frac{2}{3})$, I can plug in $-\frac{2}{3}$ for $y$ in the expression and simplify. I get $p(-\frac{2}{3})=3(-\frac{2}{3})^3-13(-\frac{2}{3})^2+11(-\frac{2}{3})+23$. This equals $-\frac{8}{9}-\frac{52}{9}-\frac{22}{3}+23$. To combine the fractions, I need a common denominator of 9. I get $-\frac{8}{9}-\frac{52}{9}-\frac{66}{9}+\frac{207}{9}$. This equals $\frac{81}{9}$. This simplifies to $9$. Therefore, the remainder is 9. 