Divide using synthetic division $$ ( 2x^{5}-x^{4}-3x^{2}-2x+1 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}2&2&-1&0&-3&-2&1\\& & 4& 6& 12& 18& \color{black}{32} \\ \hline &\color{blue}{2}&\color{blue}{3}&\color{blue}{6}&\color{blue}{9}&\color{blue}{16}&\color{orangered}{33} \end{array} $$The solution is:
$$ \frac{ 2x^{5}-x^{4}-3x^{2}-2x+1 }{ x-2 } = \color{blue}{2x^{4}+3x^{3}+6x^{2}+9x+16} ~+~ \frac{ \color{red}{ 33 } }{ x-2 } $$Explanation
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.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&2&-1&0&-3&-2&1\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}2&\color{orangered}{ 2 }&-1&0&-3&-2&1\\& & & & & & \\ \hline &\color{orangered}{2}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&2&-1&0&-3&-2&1\\& & \color{blue}{4} & & & & \\ \hline &\color{blue}{2}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 4 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrrr}2&2&\color{orangered}{ -1 }&0&-3&-2&1\\& & \color{orangered}{4} & & & & \\ \hline &2&\color{orangered}{3}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&2&-1&0&-3&-2&1\\& & 4& \color{blue}{6} & & & \\ \hline &2&\color{blue}{3}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 6 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrrr}2&2&-1&\color{orangered}{ 0 }&-3&-2&1\\& & 4& \color{orangered}{6} & & & \\ \hline &2&3&\color{orangered}{6}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 6 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&2&-1&0&-3&-2&1\\& & 4& 6& \color{blue}{12} & & \\ \hline &2&3&\color{blue}{6}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 12 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrrr}2&2&-1&0&\color{orangered}{ -3 }&-2&1\\& & 4& 6& \color{orangered}{12} & & \\ \hline &2&3&6&\color{orangered}{9}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 9 } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&2&-1&0&-3&-2&1\\& & 4& 6& 12& \color{blue}{18} & \\ \hline &2&3&6&\color{blue}{9}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 18 } = \color{orangered}{ 16 } $
$$ \begin{array}{c|rrrrrr}2&2&-1&0&-3&\color{orangered}{ -2 }&1\\& & 4& 6& 12& \color{orangered}{18} & \\ \hline &2&3&6&9&\color{orangered}{16}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 16 } = \color{blue}{ 32 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&2&-1&0&-3&-2&1\\& & 4& 6& 12& 18& \color{blue}{32} \\ \hline &2&3&6&9&\color{blue}{16}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 32 } = \color{orangered}{ 33 } $
$$ \begin{array}{c|rrrrrr}2&2&-1&0&-3&-2&\color{orangered}{ 1 }\\& & 4& 6& 12& 18& \color{orangered}{32} \\ \hline &\color{blue}{2}&\color{blue}{3}&\color{blue}{6}&\color{blue}{9}&\color{blue}{16}&\color{orangered}{33} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{4}+3x^{3}+6x^{2}+9x+16 } $ with a remainder of $ \color{red}{ 33 } $.