Synthetic division for: $\frac{60 x^{3}+43x^{2}-34x-24}{3x+2}$ If I have a polynomial to which the solutions are integers, in this case, I know how to perform the synthetic division. Also, I know how to perform the present division using long division.
But I don't know how to do it with synthetic division. I watched videos on youtube, but the polynomials presented in the shows have integer solutions.
So I would greatly appreciate If you could walk me through the solution using synthetic division. I'm sure I'll learn something new. Thank you.
$$\frac{60 x^{3}+43x^{2}-34x-24}{3x+2}$$
Edit:
The answer you're referring to has two part: One part giving the rule and I don't know how to translate it into the question in the hand; and the second part is using long division.
If the long division is the only way, please let me know and I'll accept that as an answer.
 A: In order to get a denominator of the form $x − r$, where $r$ is a root of the denominator, I would start by dividing top and bottom by the leading coefficient of the denominator, which is $3$
\begin{equation*}
\frac{60x^{3}+43x^{2}-34x-24}{3x+2}=\frac{20x^{3}+\frac{43}{3}x^{2}-\frac{34}{3}x-8}{x+\frac{2}{3}}=\frac{20x^{3}+\frac{43}{3}x^{2}-\frac{34}{3}x-8}{x-\left(-\frac{2}{3}\right)}.
\end{equation*}
Then I could use the special case of synthetic division called Ruffini's rule. Here his the computation, whose result is $20x^2+x-12$:
\begin{array}{c|rrrl}
&  x^{3} & x^{2} & x^{1} & \phantom{-} x^{0}   \\ 
& 20 & 43/3 & -34/3 & \; \; -8 &\to \text{numerator } \\
   x_0={ \color{blue}{-2/3}} & \downarrow & -40/3 & -2/3 & \;\;\phantom{-} 8 \\
              &    & {\color{gray}{=}}{\; \color{blue}{(-2/3)}}{\color{gray}{(20)}} &  {\color{gray}{=}}{\; \color{blue}{(-2/3)}}{\color{gray}{ (1)}} &  {\color{gray}{=}}{\; \color{blue}{(-2/3)}}{\color{gray}{ (-12)}} \\
    \hline  & 20 & 1 & -12 & |\phantom {-}{\color{green}0}&\to \text{quotient} \text{ and }{\color{green}{\text{remainder}}} \\
   & & { \color {gray} {=43/3-40/3 } } &  { \color {gray}{=-34/3-2/3} }  & | {   \color {gray}{=-8+8 }} \\
 &  x^{2} & x^{1} & x^{0} & |\color{green}{\text{remainder}}
    \end{array}
