Divide using synthetic division $$ ( 3x^{3}+19x^{2}-29x+10 ) \div ( 3x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&3&19&-29&10\\& & 6& 50& \color{black}{42} \\ \hline &\color{blue}{3}&\color{blue}{25}&\color{blue}{21}&\color{orangered}{52} \end{array} $$The solution is:
$$ \frac{ 3x^{3}+19x^{2}-29x+10 }{ x-2 } = \color{blue}{3x^{2}+25x+21} ~+~ \frac{ \color{red}{ 52 } }{ x-2 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -2 = 0 $ ( $ x = \color{blue}{ 2 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{2}&3&19&-29&10\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 3 }&19&-29&10\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 3 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&3&19&-29&10\\& & \color{blue}{6} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 19 } + \color{orangered}{ 6 } = \color{orangered}{ 25 } $
$$ \begin{array}{c|rrrr}2&3&\color{orangered}{ 19 }&-29&10\\& & \color{orangered}{6} & & \\ \hline &3&\color{orangered}{25}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 25 } = \color{blue}{ 50 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&3&19&-29&10\\& & 6& \color{blue}{50} & \\ \hline &3&\color{blue}{25}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -29 } + \color{orangered}{ 50 } = \color{orangered}{ 21 } $
$$ \begin{array}{c|rrrr}2&3&19&\color{orangered}{ -29 }&10\\& & 6& \color{orangered}{50} & \\ \hline &3&25&\color{orangered}{21}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 21 } = \color{blue}{ 42 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&3&19&-29&10\\& & 6& 50& \color{blue}{42} \\ \hline &3&25&\color{blue}{21}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ 42 } = \color{orangered}{ 52 } $
$$ \begin{array}{c|rrrr}2&3&19&-29&\color{orangered}{ 10 }\\& & 6& 50& \color{orangered}{42} \\ \hline &\color{blue}{3}&\color{blue}{25}&\color{blue}{21}&\color{orangered}{52} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{2}+25x+21 } $ with a remainder of $ \color{red}{ 52 } $.