Divide using synthetic division $$ ( 4x^{4}-2x^{3}+x^{2}-5x+8 ) \div ( x+2 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&4&-2&1&-5&8\\& & -8& 20& -42& \color{black}{94} \\ \hline &\color{blue}{4}&\color{blue}{-10}&\color{blue}{21}&\color{blue}{-47}&\color{orangered}{102} \end{array} $$The solution is:
$$ \dfrac{ 4x^{4}-2x^{3}+x^{2}-5x+8 }{ x+2 } = \color{blue}{4x^{3}-10x^{2}+21x-47} ~+~ \dfrac{ \color{red}{ 102 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&4&-2&1&-5&8\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 4 }&-2&1&-5&8\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 4 } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&4&-2&1&-5&8\\& & \color{blue}{-8} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrrr}-2&4&\color{orangered}{ -2 }&1&-5&8\\& & \color{orangered}{-8} & & & \\ \hline &4&\color{orangered}{-10}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&4&-2&1&-5&8\\& & -8& \color{blue}{20} & & \\ \hline &4&\color{blue}{-10}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 20 } = \color{orangered}{ 21 } $
$$ \begin{array}{c|rrrrr}-2&4&-2&\color{orangered}{ 1 }&-5&8\\& & -8& \color{orangered}{20} & & \\ \hline &4&-10&\color{orangered}{21}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 21 } = \color{blue}{ -42 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&4&-2&1&-5&8\\& & -8& 20& \color{blue}{-42} & \\ \hline &4&-10&\color{blue}{21}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ \left( -42 \right) } = \color{orangered}{ -47 } $
$$ \begin{array}{c|rrrrr}-2&4&-2&1&\color{orangered}{ -5 }&8\\& & -8& 20& \color{orangered}{-42} & \\ \hline &4&-10&21&\color{orangered}{-47}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -47 \right) } = \color{blue}{ 94 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&4&-2&1&-5&8\\& & -8& 20& -42& \color{blue}{94} \\ \hline &4&-10&21&\color{blue}{-47}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ 94 } = \color{orangered}{ 102 } $
$$ \begin{array}{c|rrrrr}-2&4&-2&1&-5&\color{orangered}{ 8 }\\& & -8& 20& -42& \color{orangered}{94} \\ \hline &\color{blue}{4}&\color{blue}{-10}&\color{blue}{21}&\color{blue}{-47}&\color{orangered}{102} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{3}-10x^{2}+21x-47 } $ with a remainder of $ \color{red}{ 102 } $.