When $6y^3+5y^2-16y+8$ is divided by $2y+3$ the quotient is $3y^2+cy+\frac{5}{2}c$. What is the remainder?
To find the remainder of a polynomial division, I can use the remainder theorem, which says that the remainder is equal to the value of the polynomial when I plug in the value of $y$ that makes the divisor zero. So in this case, I want to find $6y^3+5y^2-16y+8$ when $y=-\frac{3}{2}$, which is the root of $2y+3$. To make the calculations easier, I can use synthetic division to divide the polynomial by $y+\frac{3}{2}$. I write the coefficients of the polynomial in a row, and then write $-\frac{3}{2}$ below the first coefficient. Then I bring down the first coefficient, which is 6, and multiply it by $-\frac{3}{2}$, which gives $-9$, and write it below the second coefficient. Then I add the second coefficient and $-9$, which gives $-4$, and write it below the third coefficient. Then I multiply $-4$ by $-\frac{3}{2}$, which gives 6, and write it below the fourth coefficient.
[*, I add the fourth coefficient and 6, which gives $-10$, and write it below the fifth coefficient. Then I multiply $-10$ by $-\frac{3}{2}$, which gives 15, and write it below the sixth coefficient. Then I add the sixth coefficient and 15, which gives 23, and write it in the corner. This means that the quotient of the division is $6y^2-4y-10$, and the remainder is 23. Therefore, the answer is 23. 