Divide using synthetic division $$ ( x^{4}+3x^{3}-19x-7 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&1&3&0&-19&-7\\& & 2& 10& 20& \color{black}{2} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{10}&\color{blue}{1}&\color{orangered}{-5} \end{array} $$The solution is:
$$ \frac{ x^{4}+3x^{3}-19x-7 }{ x-2 } = \color{blue}{x^{3}+5x^{2}+10x+1} \color{red}{~-~} \frac{ \color{red}{ 5 } }{ x-2 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -2 = 0 $ ( $ x = \color{blue}{ 2 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&3&0&-19&-7\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 1 }&3&0&-19&-7\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 1 } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&3&0&-19&-7\\& & \color{blue}{2} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 2 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrr}2&1&\color{orangered}{ 3 }&0&-19&-7\\& & \color{orangered}{2} & & & \\ \hline &1&\color{orangered}{5}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 5 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&3&0&-19&-7\\& & 2& \color{blue}{10} & & \\ \hline &1&\color{blue}{5}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 10 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrr}2&1&3&\color{orangered}{ 0 }&-19&-7\\& & 2& \color{orangered}{10} & & \\ \hline &1&5&\color{orangered}{10}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 10 } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&3&0&-19&-7\\& & 2& 10& \color{blue}{20} & \\ \hline &1&5&\color{blue}{10}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -19 } + \color{orangered}{ 20 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}2&1&3&0&\color{orangered}{ -19 }&-7\\& & 2& 10& \color{orangered}{20} & \\ \hline &1&5&10&\color{orangered}{1}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 1 } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&3&0&-19&-7\\& & 2& 10& 20& \color{blue}{2} \\ \hline &1&5&10&\color{blue}{1}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 2 } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrr}2&1&3&0&-19&\color{orangered}{ -7 }\\& & 2& 10& 20& \color{orangered}{2} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{10}&\color{blue}{1}&\color{orangered}{-5} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+5x^{2}+10x+1 } $ with a remainder of $ \color{red}{ -5 } $.