Divide using synthetic division $$ ( 6x^{3}+5x^{2}-3x+3 ) \div ( -3x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&6&5&-3&3\\& & -12& 14& \color{black}{-22} \\ \hline &\color{blue}{6}&\color{blue}{-7}&\color{blue}{11}&\color{orangered}{-19} \end{array} $$The solution is:
$$ \frac{ 6x^{3}+5x^{2}-3x+3 }{ x+2 } = \color{blue}{6x^{2}-7x+11} \color{red}{~-~} \frac{ \color{red}{ 19 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&6&5&-3&3\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 6 }&5&-3&3\\& & & & \\ \hline &\color{orangered}{6}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 6 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&6&5&-3&3\\& & \color{blue}{-12} & & \\ \hline &\color{blue}{6}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrr}-2&6&\color{orangered}{ 5 }&-3&3\\& & \color{orangered}{-12} & & \\ \hline &6&\color{orangered}{-7}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ 14 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&6&5&-3&3\\& & -12& \color{blue}{14} & \\ \hline &6&\color{blue}{-7}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 14 } = \color{orangered}{ 11 } $
$$ \begin{array}{c|rrrr}-2&6&5&\color{orangered}{ -3 }&3\\& & -12& \color{orangered}{14} & \\ \hline &6&-7&\color{orangered}{11}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 11 } = \color{blue}{ -22 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&6&5&-3&3\\& & -12& 14& \color{blue}{-22} \\ \hline &6&-7&\color{blue}{11}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -22 \right) } = \color{orangered}{ -19 } $
$$ \begin{array}{c|rrrr}-2&6&5&-3&\color{orangered}{ 3 }\\& & -12& 14& \color{orangered}{-22} \\ \hline &\color{blue}{6}&\color{blue}{-7}&\color{blue}{11}&\color{orangered}{-19} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{2}-7x+11 } $ with a remainder of $ \color{red}{ -19 } $.