Answer
$\displaystyle \frac{6x^{3}+x^{2}-12x+5}{3x-4}=2x^{2}+3x+\frac{5}{3x-4}$
Work Step by Step
$ \left[\begin{array}{l}
\\\\
3x-4\ )\\
\\
\\
\\
\\
\\
\\
\\
\end{array}\right. \left.\begin{array}{lllll}
2x^{2} & 3x & & & \\
\hline 6x^{3} & +x^{2} & -12x & +5 & \\
6x^{3} & -8x^{2} & & & \\
-- & -- & & & \\
& 9x^{2} & -12x & +5 & \\
& 9x^{2} & -12x & & \\
& -- & -- & & \\
& & 0 & +5 &
\end{array}\right]$
$\displaystyle \frac{6x^{3}+x^{2}-12x+5}{3x-4}=2x^{2}+3x+\frac{5}{3x-4}$
