Divide using synthetic division $$ ( 2x^{4}-x^{3}+x^{2}+4 ) \div ( x-2 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&2&-1&1&0&4\\& & 4& 6& 14& \color{black}{28} \\ \hline &\color{blue}{2}&\color{blue}{3}&\color{blue}{7}&\color{blue}{14}&\color{orangered}{32} \end{array} $$The solution is:
$$ \dfrac{ 2x^{4}-x^{3}+x^{2}+4 }{ x-2 } = \color{blue}{2x^{3}+3x^{2}+7x+14} ~+~ \dfrac{ \color{red}{ 32 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&2&-1&1&0&4\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 2 }&-1&1&0&4\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 2 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&2&-1&1&0&4\\& & \color{blue}{4} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 4 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}2&2&\color{orangered}{ -1 }&1&0&4\\& & \color{orangered}{4} & & & \\ \hline &2&\color{orangered}{3}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 3 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&2&-1&1&0&4\\& & 4& \color{blue}{6} & & \\ \hline &2&\color{blue}{3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 6 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrrr}2&2&-1&\color{orangered}{ 1 }&0&4\\& & 4& \color{orangered}{6} & & \\ \hline &2&3&\color{orangered}{7}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 7 } = \color{blue}{ 14 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&2&-1&1&0&4\\& & 4& 6& \color{blue}{14} & \\ \hline &2&3&\color{blue}{7}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 14 } = \color{orangered}{ 14 } $
$$ \begin{array}{c|rrrrr}2&2&-1&1&\color{orangered}{ 0 }&4\\& & 4& 6& \color{orangered}{14} & \\ \hline &2&3&7&\color{orangered}{14}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 14 } = \color{blue}{ 28 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&2&-1&1&0&4\\& & 4& 6& 14& \color{blue}{28} \\ \hline &2&3&7&\color{blue}{14}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 28 } = \color{orangered}{ 32 } $
$$ \begin{array}{c|rrrrr}2&2&-1&1&0&\color{orangered}{ 4 }\\& & 4& 6& 14& \color{orangered}{28} \\ \hline &\color{blue}{2}&\color{blue}{3}&\color{blue}{7}&\color{blue}{14}&\color{orangered}{32} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}+3x^{2}+7x+14 } $ with a remainder of $ \color{red}{ 32 } $.