Divide using synthetic division $$ ( x^{3}-8x^{2}+5x+4 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&1&-8&5&4\\& & 2& -12& \color{black}{-14} \\ \hline &\color{blue}{1}&\color{blue}{-6}&\color{blue}{-7}&\color{orangered}{-10} \end{array} $$The solution is:
$$ \frac{ x^{3}-8x^{2}+5x+4 }{ x-2 } = \color{blue}{x^{2}-6x-7} \color{red}{~-~} \frac{ \color{red}{ 10 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{2}&1&-8&5&4\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 1 }&-8&5&4\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&1&-8&5&4\\& & \color{blue}{2} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 2 } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrr}2&1&\color{orangered}{ -8 }&5&4\\& & \color{orangered}{2} & & \\ \hline &1&\color{orangered}{-6}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&1&-8&5&4\\& & 2& \color{blue}{-12} & \\ \hline &1&\color{blue}{-6}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrr}2&1&-8&\color{orangered}{ 5 }&4\\& & 2& \color{orangered}{-12} & \\ \hline &1&-6&\color{orangered}{-7}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ -14 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&1&-8&5&4\\& & 2& -12& \color{blue}{-14} \\ \hline &1&-6&\color{blue}{-7}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ \left( -14 \right) } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrr}2&1&-8&5&\color{orangered}{ 4 }\\& & 2& -12& \color{orangered}{-14} \\ \hline &\color{blue}{1}&\color{blue}{-6}&\color{blue}{-7}&\color{orangered}{-10} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-6x-7 } $ with a remainder of $ \color{red}{ -10 } $.