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