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