Quotient and remainder

I brute forced the problem. ${{6a^3}\over {3a } } = 2a^2$... and so on, until I found the quotient, which is the trinomial: $\color{brown}\boxed{2a^2+5a+2}$.

When you multiply this by $3a-2$, you get: ${6a^3+11a^2-4a-4}$. The remainder to this problem is $\color{brown}\boxed{-5}$

(I'll explain it more in-depth in 45 min once I get home.)

$(6a^3+11a^2-4a-9)\over(3a-2)$

I thought about this problem as factoring. 

At the end, it will be $(3a-2)(xa^2+ya+z) + r$

In this equation, $x$, $y$, and $z$ are coeficents, and $r$ is the remainder.

We know that x must equal 2, because $2a^2(3a-2)=6a^3-4a^2$

We know that y equals 5, because (note we already have a -4 so it will add to 11): $5a(3a-2) = 15a^2-10a$

We know that z must equal 2, because(Note: we already have a -10, so it will add to -4): $2(3a-2) = 6a-4$.

When we add everything, we have: $6a^3+11a^2-4a-4$. This means that the remainder is $\color{brown}\boxed{-5}$, and the quotient is $\color{brown}\boxed{2a^2-5a+2}$