+128474 Using polynomial division, we have
x^2 - (1/3)x + (26/9)
3x - 4 [ 3x^3 - 5x^2 + 10x - 3]
3x^3 - 4x^2
------------------
-1x^2 + 10x
-1x^2 +(4/3)x
-------------------
(26/3)x - 3
(26/3)x - (104/9)
---------------------
(77/9)
So.....the answer is.......... [ x^2 - (1/3)x + (26/9) ] + [ 77/9]/[3x - 4]
-3/(-4 + 3 x) + (10 x)/(-4 + 3 x) + (-5 x^2)/(-4 + 3 x) + (3 x^3)/(-4 + 3 x)
(-3 + x (10 + x (-5 + 3 x)))/(-4 + 3 x)
(3 - 10 x + 5 x^2 - 3 x^3)/(4 - 3 x)
+128474 Using polynomial division, we have
x^2 - (1/3)x + (26/9)
3x - 4 [ 3x^3 - 5x^2 + 10x - 3]
3x^3 - 4x^2
------------------
-1x^2 + 10x
-1x^2 +(4/3)x
-------------------
(26/3)x - 3
(26/3)x - (104/9)
---------------------
(77/9)
So.....the answer is.......... [ x^2 - (1/3)x + (26/9) ] + [ 77/9]/[3x - 4]
+26367 (3x^3-5x^2+10x-3)/(3x-4)
In mathematics, Horner's method (also known as Horner scheme in the UK or Horner's rule in the U.S.) an algorithm for calculating polynomials
see: https://en.wikipedia.org/wiki/Horner%27s_method#cite_note-HornerRule-2
$$\small{
\begin{array}{lrrl}
& f_1{(x)} &=& 3x^3-5x^2+10x-3\\
& f_2{(x)} &=& 3x-4\\
&\text{Divide } f_1{(x)} \text{ by } f_2{(x)} \text{ using Honer's method }
\end{array}
}$$
$$\small{\text{$
\begin{array}{r|crcrcrc|r}
3 &\quad& 3 &\quad& -5 &\quad& 10 &\quad& -3 \\
\hline
&&&&&&&\\
4 &\quad& &\quad& 4 &\quad& -\dfrac{4}{3} &\quad& 4\cdot \dfrac{26}{9}\\
\hline
&&&&&&&\\
&\quad& 1 &\quad& -1\cdot \dfrac{1}{3} &\quad&\dfrac{26}{3}\cdot \dfrac{1}{3} &\quad & \dfrac{77}{9}
\end{array}
$}}$$
$$1\cdot x^2 - \dfrac{1}{3}\cdot x + \dfrac{26}{9}
+ \dfrac{77}{9(3x-4)}$$
