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