Divide using synthetic division $$ ( 6x^{5}-4x^{3}+3x^{2}-7x+5 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}2&6&0&-4&3&-7&5\\& & 12& 24& 40& 86& \color{black}{158} \\ \hline &\color{blue}{6}&\color{blue}{12}&\color{blue}{20}&\color{blue}{43}&\color{blue}{79}&\color{orangered}{163} \end{array} $$The solution is:
$$ \frac{ 6x^{5}-4x^{3}+3x^{2}-7x+5 }{ x-2 } = \color{blue}{6x^{4}+12x^{3}+20x^{2}+43x+79} ~+~ \frac{ \color{red}{ 163 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&6&0&-4&3&-7&5\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}2&\color{orangered}{ 6 }&0&-4&3&-7&5\\& & & & & & \\ \hline &\color{orangered}{6}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 6 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&6&0&-4&3&-7&5\\& & \color{blue}{12} & & & & \\ \hline &\color{blue}{6}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 12 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrrrr}2&6&\color{orangered}{ 0 }&-4&3&-7&5\\& & \color{orangered}{12} & & & & \\ \hline &6&\color{orangered}{12}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 12 } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&6&0&-4&3&-7&5\\& & 12& \color{blue}{24} & & & \\ \hline &6&\color{blue}{12}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 24 } = \color{orangered}{ 20 } $
$$ \begin{array}{c|rrrrrr}2&6&0&\color{orangered}{ -4 }&3&-7&5\\& & 12& \color{orangered}{24} & & & \\ \hline &6&12&\color{orangered}{20}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 20 } = \color{blue}{ 40 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&6&0&-4&3&-7&5\\& & 12& 24& \color{blue}{40} & & \\ \hline &6&12&\color{blue}{20}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 40 } = \color{orangered}{ 43 } $
$$ \begin{array}{c|rrrrrr}2&6&0&-4&\color{orangered}{ 3 }&-7&5\\& & 12& 24& \color{orangered}{40} & & \\ \hline &6&12&20&\color{orangered}{43}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 43 } = \color{blue}{ 86 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&6&0&-4&3&-7&5\\& & 12& 24& 40& \color{blue}{86} & \\ \hline &6&12&20&\color{blue}{43}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 86 } = \color{orangered}{ 79 } $
$$ \begin{array}{c|rrrrrr}2&6&0&-4&3&\color{orangered}{ -7 }&5\\& & 12& 24& 40& \color{orangered}{86} & \\ \hline &6&12&20&43&\color{orangered}{79}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 79 } = \color{blue}{ 158 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&6&0&-4&3&-7&5\\& & 12& 24& 40& 86& \color{blue}{158} \\ \hline &6&12&20&43&\color{blue}{79}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 158 } = \color{orangered}{ 163 } $
$$ \begin{array}{c|rrrrrr}2&6&0&-4&3&-7&\color{orangered}{ 5 }\\& & 12& 24& 40& 86& \color{orangered}{158} \\ \hline &\color{blue}{6}&\color{blue}{12}&\color{blue}{20}&\color{blue}{43}&\color{blue}{79}&\color{orangered}{163} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{4}+12x^{3}+20x^{2}+43x+79 } $ with a remainder of $ \color{red}{ 163 } $.