Divide using synthetic division $$ ( 4x^{3}+6x^{2}+3x-18 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&4&6&3&-18\\& & -12& 18& \color{black}{-63} \\ \hline &\color{blue}{4}&\color{blue}{-6}&\color{blue}{21}&\color{orangered}{-81} \end{array} $$The solution is:
$$ \frac{ 4x^{3}+6x^{2}+3x-18 }{ x+3 } = \color{blue}{4x^{2}-6x+21} \color{red}{~-~} \frac{ \color{red}{ 81 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&4&6&3&-18\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 4 }&6&3&-18\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 4 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&4&6&3&-18\\& & \color{blue}{-12} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrr}-3&4&\color{orangered}{ 6 }&3&-18\\& & \color{orangered}{-12} & & \\ \hline &4&\color{orangered}{-6}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&4&6&3&-18\\& & -12& \color{blue}{18} & \\ \hline &4&\color{blue}{-6}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 18 } = \color{orangered}{ 21 } $
$$ \begin{array}{c|rrrr}-3&4&6&\color{orangered}{ 3 }&-18\\& & -12& \color{orangered}{18} & \\ \hline &4&-6&\color{orangered}{21}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 21 } = \color{blue}{ -63 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&4&6&3&-18\\& & -12& 18& \color{blue}{-63} \\ \hline &4&-6&\color{blue}{21}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -18 } + \color{orangered}{ \left( -63 \right) } = \color{orangered}{ -81 } $
$$ \begin{array}{c|rrrr}-3&4&6&3&\color{orangered}{ -18 }\\& & -12& 18& \color{orangered}{-63} \\ \hline &\color{blue}{4}&\color{blue}{-6}&\color{blue}{21}&\color{orangered}{-81} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}-6x+21 } $ with a remainder of $ \color{red}{ -81 } $.