Divide using synthetic division $$ ( 5x^{5}-2x^{3}+3x^{2}-2x+2 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}2&5&0&-2&3&-2&2\\& & 10& 20& 36& 78& \color{black}{152} \\ \hline &\color{blue}{5}&\color{blue}{10}&\color{blue}{18}&\color{blue}{39}&\color{blue}{76}&\color{orangered}{154} \end{array} $$The solution is:
$$ \frac{ 5x^{5}-2x^{3}+3x^{2}-2x+2 }{ x-2 } = \color{blue}{5x^{4}+10x^{3}+18x^{2}+39x+76} ~+~ \frac{ \color{red}{ 154 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&5&0&-2&3&-2&2\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}2&\color{orangered}{ 5 }&0&-2&3&-2&2\\& & & & & & \\ \hline &\color{orangered}{5}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 5 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&5&0&-2&3&-2&2\\& & \color{blue}{10} & & & & \\ \hline &\color{blue}{5}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 10 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrrr}2&5&\color{orangered}{ 0 }&-2&3&-2&2\\& & \color{orangered}{10} & & & & \\ \hline &5&\color{orangered}{10}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 10 } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&5&0&-2&3&-2&2\\& & 10& \color{blue}{20} & & & \\ \hline &5&\color{blue}{10}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 20 } = \color{orangered}{ 18 } $
$$ \begin{array}{c|rrrrrr}2&5&0&\color{orangered}{ -2 }&3&-2&2\\& & 10& \color{orangered}{20} & & & \\ \hline &5&10&\color{orangered}{18}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 18 } = \color{blue}{ 36 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&5&0&-2&3&-2&2\\& & 10& 20& \color{blue}{36} & & \\ \hline &5&10&\color{blue}{18}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 36 } = \color{orangered}{ 39 } $
$$ \begin{array}{c|rrrrrr}2&5&0&-2&\color{orangered}{ 3 }&-2&2\\& & 10& 20& \color{orangered}{36} & & \\ \hline &5&10&18&\color{orangered}{39}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 39 } = \color{blue}{ 78 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&5&0&-2&3&-2&2\\& & 10& 20& 36& \color{blue}{78} & \\ \hline &5&10&18&\color{blue}{39}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 78 } = \color{orangered}{ 76 } $
$$ \begin{array}{c|rrrrrr}2&5&0&-2&3&\color{orangered}{ -2 }&2\\& & 10& 20& 36& \color{orangered}{78} & \\ \hline &5&10&18&39&\color{orangered}{76}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 76 } = \color{blue}{ 152 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&5&0&-2&3&-2&2\\& & 10& 20& 36& 78& \color{blue}{152} \\ \hline &5&10&18&39&\color{blue}{76}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 152 } = \color{orangered}{ 154 } $
$$ \begin{array}{c|rrrrrr}2&5&0&-2&3&-2&\color{orangered}{ 2 }\\& & 10& 20& 36& 78& \color{orangered}{152} \\ \hline &\color{blue}{5}&\color{blue}{10}&\color{blue}{18}&\color{blue}{39}&\color{blue}{76}&\color{orangered}{154} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{4}+10x^{3}+18x^{2}+39x+76 } $ with a remainder of $ \color{red}{ 154 } $.