What is the remainder when $6x^3-15x^2+21x-23$ is divided by $3x-6$?
To find the remainder when a polynomial is divided by another polynomial, I can use the remainder theorem, which says that the remainder is equal to the value of the dividend polynomial when I plug in the value of $x$ that makes the divisor zero. So in this case, I want to find the value of $x$ that makes $3x-6=0$. I can do this by adding $6$ to both sides and then dividing by $3$, which gives me $x=2$. Now I just need to plug in $x=2$ into the dividend polynomial and simplify. I get $6(2)^3-15(2)^2+21(2)-23$, which is equal to $6(8)-15(4)+42-23$. This is equal to $48-60+42-23$, which is equal to $7$. Therefore, the remainder is $7$. 