#1**+3 **

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.)

BuilderBoi Feb 2, 2022

#2**+3 **

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

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 \(\color{brown}\boxed{2a^2-5a+2}\)

BuilderBoi Feb 2, 2022