Divide using synthetic division $$ ( x^{5}-12x^{4}+19x^{3}+22x^{2}+2x+14 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}3&1&-12&19&22&2&14\\& & 3& -27& -24& -6& \color{black}{-12} \\ \hline &\color{blue}{1}&\color{blue}{-9}&\color{blue}{-8}&\color{blue}{-2}&\color{blue}{-4}&\color{orangered}{2} \end{array} $$The solution is:
$$ \frac{ x^{5}-12x^{4}+19x^{3}+22x^{2}+2x+14 }{ x-3 } = \color{blue}{x^{4}-9x^{3}-8x^{2}-2x-4} ~+~ \frac{ \color{red}{ 2 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&-12&19&22&2&14\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}3&\color{orangered}{ 1 }&-12&19&22&2&14\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&-12&19&22&2&14\\& & \color{blue}{3} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 3 } = \color{orangered}{ -9 } $
$$ \begin{array}{c|rrrrrr}3&1&\color{orangered}{ -12 }&19&22&2&14\\& & \color{orangered}{3} & & & & \\ \hline &1&\color{orangered}{-9}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&-12&19&22&2&14\\& & 3& \color{blue}{-27} & & & \\ \hline &1&\color{blue}{-9}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 19 } + \color{orangered}{ \left( -27 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrrr}3&1&-12&\color{orangered}{ 19 }&22&2&14\\& & 3& \color{orangered}{-27} & & & \\ \hline &1&-9&\color{orangered}{-8}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ -24 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&-12&19&22&2&14\\& & 3& -27& \color{blue}{-24} & & \\ \hline &1&-9&\color{blue}{-8}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 22 } + \color{orangered}{ \left( -24 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrrr}3&1&-12&19&\color{orangered}{ 22 }&2&14\\& & 3& -27& \color{orangered}{-24} & & \\ \hline &1&-9&-8&\color{orangered}{-2}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&-12&19&22&2&14\\& & 3& -27& -24& \color{blue}{-6} & \\ \hline &1&-9&-8&\color{blue}{-2}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrrr}3&1&-12&19&22&\color{orangered}{ 2 }&14\\& & 3& -27& -24& \color{orangered}{-6} & \\ \hline &1&-9&-8&-2&\color{orangered}{-4}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&-12&19&22&2&14\\& & 3& -27& -24& -6& \color{blue}{-12} \\ \hline &1&-9&-8&-2&\color{blue}{-4}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 14 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrrr}3&1&-12&19&22&2&\color{orangered}{ 14 }\\& & 3& -27& -24& -6& \color{orangered}{-12} \\ \hline &\color{blue}{1}&\color{blue}{-9}&\color{blue}{-8}&\color{blue}{-2}&\color{blue}{-4}&\color{orangered}{2} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}-9x^{3}-8x^{2}-2x-4 } $ with a remainder of $ \color{red}{ 2 } $.