Divide using synthetic division $$ ( 2x^{3}-4x^{2}+3x-8 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&2&-4&3&-8\\& & -6& 30& \color{black}{-99} \\ \hline &\color{blue}{2}&\color{blue}{-10}&\color{blue}{33}&\color{orangered}{-107} \end{array} $$The solution is:
$$ \frac{ 2x^{3}-4x^{2}+3x-8 }{ x+3 } = \color{blue}{2x^{2}-10x+33} \color{red}{~-~} \frac{ \color{red}{ 107 } }{ x+3 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 3 = 0 $ ( $ x = \color{blue}{ -3 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&2&-4&3&-8\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 2 }&-4&3&-8\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 2 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&2&-4&3&-8\\& & \color{blue}{-6} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrr}-3&2&\color{orangered}{ -4 }&3&-8\\& & \color{orangered}{-6} & & \\ \hline &2&\color{orangered}{-10}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&2&-4&3&-8\\& & -6& \color{blue}{30} & \\ \hline &2&\color{blue}{-10}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 30 } = \color{orangered}{ 33 } $
$$ \begin{array}{c|rrrr}-3&2&-4&\color{orangered}{ 3 }&-8\\& & -6& \color{orangered}{30} & \\ \hline &2&-10&\color{orangered}{33}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 33 } = \color{blue}{ -99 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&2&-4&3&-8\\& & -6& 30& \color{blue}{-99} \\ \hline &2&-10&\color{blue}{33}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ \left( -99 \right) } = \color{orangered}{ -107 } $
$$ \begin{array}{c|rrrr}-3&2&-4&3&\color{orangered}{ -8 }\\& & -6& 30& \color{orangered}{-99} \\ \hline &\color{blue}{2}&\color{blue}{-10}&\color{blue}{33}&\color{orangered}{-107} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}-10x+33 } $ with a remainder of $ \color{red}{ -107 } $.