Divide using synthetic division $$ ( x^{3}-6x^{2}+12x-3 ) \div ( -x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&1&-6&12&-3\\& & -2& 16& \color{black}{-56} \\ \hline &\color{blue}{1}&\color{blue}{-8}&\color{blue}{28}&\color{orangered}{-59} \end{array} $$The solution is:
$$ \frac{ x^{3}-6x^{2}+12x-3 }{ x+2 } = \color{blue}{x^{2}-8x+28} \color{red}{~-~} \frac{ \color{red}{ 59 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&-6&12&-3\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 1 }&-6&12&-3\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&-6&12&-3\\& & \color{blue}{-2} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrr}-2&1&\color{orangered}{ -6 }&12&-3\\& & \color{orangered}{-2} & & \\ \hline &1&\color{orangered}{-8}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&-6&12&-3\\& & -2& \color{blue}{16} & \\ \hline &1&\color{blue}{-8}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ 16 } = \color{orangered}{ 28 } $
$$ \begin{array}{c|rrrr}-2&1&-6&\color{orangered}{ 12 }&-3\\& & -2& \color{orangered}{16} & \\ \hline &1&-8&\color{orangered}{28}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 28 } = \color{blue}{ -56 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&-6&12&-3\\& & -2& 16& \color{blue}{-56} \\ \hline &1&-8&\color{blue}{28}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ \left( -56 \right) } = \color{orangered}{ -59 } $
$$ \begin{array}{c|rrrr}-2&1&-6&12&\color{orangered}{ -3 }\\& & -2& 16& \color{orangered}{-56} \\ \hline &\color{blue}{1}&\color{blue}{-8}&\color{blue}{28}&\color{orangered}{-59} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-8x+28 } $ with a remainder of $ \color{red}{ -59 } $.