Divide using synthetic division $$ ( x^{3}+3x^{2}-3x+4 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&1&3&-3&4\\& & 3& 18& \color{black}{45} \\ \hline &\color{blue}{1}&\color{blue}{6}&\color{blue}{15}&\color{orangered}{49} \end{array} $$The solution is:
$$ \frac{ x^{3}+3x^{2}-3x+4 }{ x-3 } = \color{blue}{x^{2}+6x+15} ~+~ \frac{ \color{red}{ 49 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&3&-3&4\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 1 }&3&-3&4\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 1 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&3&-3&4\\& & \color{blue}{3} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 3 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrr}3&1&\color{orangered}{ 3 }&-3&4\\& & \color{orangered}{3} & & \\ \hline &1&\color{orangered}{6}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 6 } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&3&-3&4\\& & 3& \color{blue}{18} & \\ \hline &1&\color{blue}{6}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 18 } = \color{orangered}{ 15 } $
$$ \begin{array}{c|rrrr}3&1&3&\color{orangered}{ -3 }&4\\& & 3& \color{orangered}{18} & \\ \hline &1&6&\color{orangered}{15}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 15 } = \color{blue}{ 45 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&3&-3&4\\& & 3& 18& \color{blue}{45} \\ \hline &1&6&\color{blue}{15}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 45 } = \color{orangered}{ 49 } $
$$ \begin{array}{c|rrrr}3&1&3&-3&\color{orangered}{ 4 }\\& & 3& 18& \color{orangered}{45} \\ \hline &\color{blue}{1}&\color{blue}{6}&\color{blue}{15}&\color{orangered}{49} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}+6x+15 } $ with a remainder of $ \color{red}{ 49 } $.