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