Divide using synthetic division $$ ( 4x^{4}+3x^{3}+3 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&4&3&0&0&3\\& & -12& 27& -81& \color{black}{243} \\ \hline &\color{blue}{4}&\color{blue}{-9}&\color{blue}{27}&\color{blue}{-81}&\color{orangered}{246} \end{array} $$The solution is:
$$ \frac{ 4x^{4}+3x^{3}+3 }{ x+3 } = \color{blue}{4x^{3}-9x^{2}+27x-81} ~+~ \frac{ \color{red}{ 246 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&3&0&0&3\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 4 }&3&0&0&3\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&3&0&0&3\\& & \color{blue}{-12} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -9 } $
$$ \begin{array}{c|rrrrr}-3&4&\color{orangered}{ 3 }&0&0&3\\& & \color{orangered}{-12} & & & \\ \hline &4&\color{orangered}{-9}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -9 \right) } = \color{blue}{ 27 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&3&0&0&3\\& & -12& \color{blue}{27} & & \\ \hline &4&\color{blue}{-9}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 27 } = \color{orangered}{ 27 } $
$$ \begin{array}{c|rrrrr}-3&4&3&\color{orangered}{ 0 }&0&3\\& & -12& \color{orangered}{27} & & \\ \hline &4&-9&\color{orangered}{27}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 27 } = \color{blue}{ -81 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&3&0&0&3\\& & -12& 27& \color{blue}{-81} & \\ \hline &4&-9&\color{blue}{27}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -81 \right) } = \color{orangered}{ -81 } $
$$ \begin{array}{c|rrrrr}-3&4&3&0&\color{orangered}{ 0 }&3\\& & -12& 27& \color{orangered}{-81} & \\ \hline &4&-9&27&\color{orangered}{-81}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -81 \right) } = \color{blue}{ 243 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&3&0&0&3\\& & -12& 27& -81& \color{blue}{243} \\ \hline &4&-9&27&\color{blue}{-81}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 243 } = \color{orangered}{ 246 } $
$$ \begin{array}{c|rrrrr}-3&4&3&0&0&\color{orangered}{ 3 }\\& & -12& 27& -81& \color{orangered}{243} \\ \hline &\color{blue}{4}&\color{blue}{-9}&\color{blue}{27}&\color{blue}{-81}&\color{orangered}{246} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{3}-9x^{2}+27x-81 } $ with a remainder of $ \color{red}{ 246 } $.