When $3z^3-4z^2-14z+3$ is divided by $3z+5$, the quotient is $z^2-3z+\frac{1}{3}$. What is the remainder?
I know that in polynomial division, the remainder is the part that is left over after I subtract the product of the divisor and the quotient from the dividend. So, to find the remainder, I need to multiply $3z+5$ by $z^2-3Z+\frac{1}{3}$ and then subtract that from $3z^3-4z^2-14z+3$. To multiply two polynomials, I can use the distributive property and then combine like terms. So, $3z+5$ times $z^2-3z+\frac{1}{3}$ is $3z(z^2-3z+\frac{1}{3})+5(z^2-3z+\frac{1}{3})$. Expanding the parentheses, I get $3z^3-9z^2+z+5z^2-15z+\frac{5}{3}$. Combining like terms, I get $3z^3-4z^2-14z+\frac{5}{3}$. Now, to find the remainder, I need to subtract this from $3z^3-4z^2-14z+3$. To subtract two polynomials, I can align them vertically and subtract the coefficients of each term. So, I have $3z^3-4z^2-14z+3$ minus $3z^3-4z^2-14z+\frac{5}{3}$. Subtracting the coefficients, I get $0z^3+0z^2+0z+\frac{4}{3}$. This simplifies to $\frac{4}{3}$. Therefore, the remainder is $\frac{4}{3}$. 