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