Answer
$ 4a^2 +27a+35$.
Work Step by Step
The given expression is
$\Rightarrow (12a^3+65a^2-3a-140)\div(3a-4)$
$\begin{matrix}
& 4a^2 & +27a &+35 & & \leftarrow &Quotient\\
&-- &-- &--&--& \\
3a-4) &12a^3&+65a^2&-3a&-140 & \\
& 12a^3 & -16a^2 & & & \leftarrow &4a^2(3a-4) \\
& -- & -- & & & \leftarrow &subtract \\
& 0 & +81a^2 & -3a & & \\
& & 81a^2 & -108a & & \leftarrow & 27a(3a-4) \\
& & -- & -- & & \leftarrow & subtract \\
& & 0&105a &-140 & \\
& & & 105a& -140 & \leftarrow & 35(3a-4) \\
& & & -- & -- & \leftarrow & subtract \\
& & & 0 & 0 & \leftarrow & Remainder
\end{matrix}$
The answer is
$\Rightarrow Quotient + \frac{Remainder}{Divisor}$
$\Rightarrow 4a^2 +27a+35+\frac{0}{3a-4}$
Simplify.
$\Rightarrow 4a^2 +27a+35$.
