Divide using synthetic division $$ ( x^{5}+2x^{4}-41x^{3}+21x^{2}+44x+12 ) \div ( x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}5&1&2&-41&21&44&12\\& & 5& 35& -30& -45& \color{black}{-5} \\ \hline &\color{blue}{1}&\color{blue}{7}&\color{blue}{-6}&\color{blue}{-9}&\color{blue}{-1}&\color{orangered}{7} \end{array} $$The solution is:
$$ \frac{ x^{5}+2x^{4}-41x^{3}+21x^{2}+44x+12 }{ x-5 } = \color{blue}{x^{4}+7x^{3}-6x^{2}-9x-1} ~+~ \frac{ \color{red}{ 7 } }{ x-5 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -5 = 0 $ ( $ x = \color{blue}{ 5 } $ ) at the left.
$$ \begin{array}{c|rrrrrr}\color{blue}{5}&1&2&-41&21&44&12\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}5&\color{orangered}{ 1 }&2&-41&21&44&12\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 1 } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{5}&1&2&-41&21&44&12\\& & \color{blue}{5} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 5 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrrrr}5&1&\color{orangered}{ 2 }&-41&21&44&12\\& & \color{orangered}{5} & & & & \\ \hline &1&\color{orangered}{7}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 7 } = \color{blue}{ 35 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{5}&1&2&-41&21&44&12\\& & 5& \color{blue}{35} & & & \\ \hline &1&\color{blue}{7}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -41 } + \color{orangered}{ 35 } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrrr}5&1&2&\color{orangered}{ -41 }&21&44&12\\& & 5& \color{orangered}{35} & & & \\ \hline &1&7&\color{orangered}{-6}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ -30 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{5}&1&2&-41&21&44&12\\& & 5& 35& \color{blue}{-30} & & \\ \hline &1&7&\color{blue}{-6}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 21 } + \color{orangered}{ \left( -30 \right) } = \color{orangered}{ -9 } $
$$ \begin{array}{c|rrrrrr}5&1&2&-41&\color{orangered}{ 21 }&44&12\\& & 5& 35& \color{orangered}{-30} & & \\ \hline &1&7&-6&\color{orangered}{-9}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -9 \right) } = \color{blue}{ -45 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{5}&1&2&-41&21&44&12\\& & 5& 35& -30& \color{blue}{-45} & \\ \hline &1&7&-6&\color{blue}{-9}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 44 } + \color{orangered}{ \left( -45 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrrr}5&1&2&-41&21&\color{orangered}{ 44 }&12\\& & 5& 35& -30& \color{orangered}{-45} & \\ \hline &1&7&-6&-9&\color{orangered}{-1}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ -5 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{5}&1&2&-41&21&44&12\\& & 5& 35& -30& -45& \color{blue}{-5} \\ \hline &1&7&-6&-9&\color{blue}{-1}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrrrr}5&1&2&-41&21&44&\color{orangered}{ 12 }\\& & 5& 35& -30& -45& \color{orangered}{-5} \\ \hline &\color{blue}{1}&\color{blue}{7}&\color{blue}{-6}&\color{blue}{-9}&\color{blue}{-1}&\color{orangered}{7} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}+7x^{3}-6x^{2}-9x-1 } $ with a remainder of $ \color{red}{ 7 } $.