Divide using synthetic division $$ ( 2x^{3}-6x^{2}+8x ) \div ( x+2 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&2&-6&8&0\\& & -4& 20& \color{black}{-56} \\ \hline &\color{blue}{2}&\color{blue}{-10}&\color{blue}{28}&\color{orangered}{-56} \end{array} $$The solution is:
$$ \dfrac{ 2x^{3}-6x^{2}+8x }{ x+2 } = \color{blue}{2x^{2}-10x+28} \color{red}{~-~} \dfrac{ \color{red}{ 56 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&2&-6&8&0\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 2 }&-6&8&0\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 2 } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&2&-6&8&0\\& & \color{blue}{-4} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrr}-2&2&\color{orangered}{ -6 }&8&0\\& & \color{orangered}{-4} & & \\ \hline &2&\color{orangered}{-10}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&2&-6&8&0\\& & -4& \color{blue}{20} & \\ \hline &2&\color{blue}{-10}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ 20 } = \color{orangered}{ 28 } $
$$ \begin{array}{c|rrrr}-2&2&-6&\color{orangered}{ 8 }&0\\& & -4& \color{orangered}{20} & \\ \hline &2&-10&\color{orangered}{28}& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&2&-6&8&0\\& & -4& 20& \color{blue}{-56} \\ \hline &2&-10&\color{blue}{28}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -56 \right) } = \color{orangered}{ -56 } $
$$ \begin{array}{c|rrrr}-2&2&-6&8&\color{orangered}{ 0 }\\& & -4& 20& \color{orangered}{-56} \\ \hline &\color{blue}{2}&\color{blue}{-10}&\color{blue}{28}&\color{orangered}{-56} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}-10x+28 } $ with a remainder of $ \color{red}{ -56 } $.