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