Divide using synthetic division $$ ( 6x^{4}-13x^{3}+4x-1 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&6&-13&0&4&-1\\& & 12& -2& -4& \color{black}{0} \\ \hline &\color{blue}{6}&\color{blue}{-1}&\color{blue}{-2}&\color{blue}{0}&\color{orangered}{-1} \end{array} $$The solution is:
$$ \frac{ 6x^{4}-13x^{3}+4x-1 }{ x-2 } = \color{blue}{6x^{3}-x^{2}-2x} \color{red}{~-~} \frac{ \color{red}{ 1 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&6&-13&0&4&-1\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 6 }&-13&0&4&-1\\& & & & & \\ \hline &\color{orangered}{6}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&6&-13&0&4&-1\\& & \color{blue}{12} & & & \\ \hline &\color{blue}{6}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -13 } + \color{orangered}{ 12 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrr}2&6&\color{orangered}{ -13 }&0&4&-1\\& & \color{orangered}{12} & & & \\ \hline &6&\color{orangered}{-1}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&6&-13&0&4&-1\\& & 12& \color{blue}{-2} & & \\ \hline &6&\color{blue}{-1}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}2&6&-13&\color{orangered}{ 0 }&4&-1\\& & 12& \color{orangered}{-2} & & \\ \hline &6&-1&\color{orangered}{-2}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&6&-13&0&4&-1\\& & 12& -2& \color{blue}{-4} & \\ \hline &6&-1&\color{blue}{-2}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}2&6&-13&0&\color{orangered}{ 4 }&-1\\& & 12& -2& \color{orangered}{-4} & \\ \hline &6&-1&-2&\color{orangered}{0}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&6&-13&0&4&-1\\& & 12& -2& -4& \color{blue}{0} \\ \hline &6&-1&-2&\color{blue}{0}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 0 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrr}2&6&-13&0&4&\color{orangered}{ -1 }\\& & 12& -2& -4& \color{orangered}{0} \\ \hline &\color{blue}{6}&\color{blue}{-1}&\color{blue}{-2}&\color{blue}{0}&\color{orangered}{-1} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{3}-x^{2}-2x } $ with a remainder of $ \color{red}{ -1 } $.