Divide using synthetic division $$ ( x^{2}-6x+3 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrr}-2&1&-6&3\\& & -2& \color{black}{16} \\ \hline &\color{blue}{1}&\color{blue}{-8}&\color{orangered}{19} \end{array} $$The solution is:
$$ \frac{ x^{2}-6x+3 }{ x+2 } = \color{blue}{x-8} ~+~ \frac{ \color{red}{ 19 } }{ x+2 } $$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.
$$ \begin{array}{c|rrr}\color{blue}{-2}&1&-6&3\\& & & \\ \hline &&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrr}-2&\color{orangered}{ 1 }&-6&3\\& & & \\ \hline &\color{orangered}{1}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrr}\color{blue}{-2}&1&-6&3\\& & \color{blue}{-2} & \\ \hline &\color{blue}{1}&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrr}-2&1&\color{orangered}{ -6 }&3\\& & \color{orangered}{-2} & \\ \hline &1&\color{orangered}{-8}& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrr}\color{blue}{-2}&1&-6&3\\& & -2& \color{blue}{16} \\ \hline &1&\color{blue}{-8}& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 16 } = \color{orangered}{ 19 } $
$$ \begin{array}{c|rrr}-2&1&-6&\color{orangered}{ 3 }\\& & -2& \color{orangered}{16} \\ \hline &\color{blue}{1}&\color{blue}{-8}&\color{orangered}{19} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x-8 } $ with a remainder of $ \color{red}{ 19 } $.