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