Divide using synthetic division $$ ( x^{5}-3x^{4}+12x-3 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}3&1&-3&0&0&12&-3\\& & 3& 0& 0& 0& \color{black}{36} \\ \hline &\color{blue}{1}&\color{blue}{0}&\color{blue}{0}&\color{blue}{0}&\color{blue}{12}&\color{orangered}{33} \end{array} $$The solution is:
$$ \frac{ x^{5}-3x^{4}+12x-3 }{ x-3 } = \color{blue}{x^{4}+12} ~+~ \frac{ \color{red}{ 33 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&-3&0&0&12&-3\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}3&\color{orangered}{ 1 }&-3&0&0&12&-3\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&-3&0&0&12&-3\\& & \color{blue}{3} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 3 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrr}3&1&\color{orangered}{ -3 }&0&0&12&-3\\& & \color{orangered}{3} & & & & \\ \hline &1&\color{orangered}{0}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&-3&0&0&12&-3\\& & 3& \color{blue}{0} & & & \\ \hline &1&\color{blue}{0}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrr}3&1&-3&\color{orangered}{ 0 }&0&12&-3\\& & 3& \color{orangered}{0} & & & \\ \hline &1&0&\color{orangered}{0}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&-3&0&0&12&-3\\& & 3& 0& \color{blue}{0} & & \\ \hline &1&0&\color{blue}{0}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrr}3&1&-3&0&\color{orangered}{ 0 }&12&-3\\& & 3& 0& \color{orangered}{0} & & \\ \hline &1&0&0&\color{orangered}{0}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&-3&0&0&12&-3\\& & 3& 0& 0& \color{blue}{0} & \\ \hline &1&0&0&\color{blue}{0}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ 0 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrrrr}3&1&-3&0&0&\color{orangered}{ 12 }&-3\\& & 3& 0& 0& \color{orangered}{0} & \\ \hline &1&0&0&0&\color{orangered}{12}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 12 } = \color{blue}{ 36 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&-3&0&0&12&-3\\& & 3& 0& 0& 0& \color{blue}{36} \\ \hline &1&0&0&0&\color{blue}{12}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 36 } = \color{orangered}{ 33 } $
$$ \begin{array}{c|rrrrrr}3&1&-3&0&0&12&\color{orangered}{ -3 }\\& & 3& 0& 0& 0& \color{orangered}{36} \\ \hline &\color{blue}{1}&\color{blue}{0}&\color{blue}{0}&\color{blue}{0}&\color{blue}{12}&\color{orangered}{33} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}+12 } $ with a remainder of $ \color{red}{ 33 } $.