Divide using synthetic division $$ ( x^{5}-15x^{2}-18x-12 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}3&1&0&0&-15&-18&-12\\& & 3& 9& 27& 36& \color{black}{54} \\ \hline &\color{blue}{1}&\color{blue}{3}&\color{blue}{9}&\color{blue}{12}&\color{blue}{18}&\color{orangered}{42} \end{array} $$The solution is:
$$ \frac{ x^{5}-15x^{2}-18x-12 }{ x-3 } = \color{blue}{x^{4}+3x^{3}+9x^{2}+12x+18} ~+~ \frac{ \color{red}{ 42 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&0&0&-15&-18&-12\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}3&\color{orangered}{ 1 }&0&0&-15&-18&-12\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&0&0&-15&-18&-12\\& & \color{blue}{3} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 3 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrrr}3&1&\color{orangered}{ 0 }&0&-15&-18&-12\\& & \color{orangered}{3} & & & & \\ \hline &1&\color{orangered}{3}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 3 } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&0&0&-15&-18&-12\\& & 3& \color{blue}{9} & & & \\ \hline &1&\color{blue}{3}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 9 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrrr}3&1&0&\color{orangered}{ 0 }&-15&-18&-12\\& & 3& \color{orangered}{9} & & & \\ \hline &1&3&\color{orangered}{9}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 9 } = \color{blue}{ 27 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&0&0&-15&-18&-12\\& & 3& 9& \color{blue}{27} & & \\ \hline &1&3&\color{blue}{9}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ 27 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrrrr}3&1&0&0&\color{orangered}{ -15 }&-18&-12\\& & 3& 9& \color{orangered}{27} & & \\ \hline &1&3&9&\color{orangered}{12}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 12 } = \color{blue}{ 36 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&0&0&-15&-18&-12\\& & 3& 9& 27& \color{blue}{36} & \\ \hline &1&3&9&\color{blue}{12}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -18 } + \color{orangered}{ 36 } = \color{orangered}{ 18 } $
$$ \begin{array}{c|rrrrrr}3&1&0&0&-15&\color{orangered}{ -18 }&-12\\& & 3& 9& 27& \color{orangered}{36} & \\ \hline &1&3&9&12&\color{orangered}{18}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 18 } = \color{blue}{ 54 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&0&0&-15&-18&-12\\& & 3& 9& 27& 36& \color{blue}{54} \\ \hline &1&3&9&12&\color{blue}{18}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 54 } = \color{orangered}{ 42 } $
$$ \begin{array}{c|rrrrrr}3&1&0&0&-15&-18&\color{orangered}{ -12 }\\& & 3& 9& 27& 36& \color{orangered}{54} \\ \hline &\color{blue}{1}&\color{blue}{3}&\color{blue}{9}&\color{blue}{12}&\color{blue}{18}&\color{orangered}{42} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}+3x^{3}+9x^{2}+12x+18 } $ with a remainder of $ \color{red}{ 42 } $.