Divide using synthetic division $$ ( x^{4}+3x^{3}+x^{2}+3x-2 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&1&3&1&3&-2\\& & 2& 10& 22& \color{black}{50} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{11}&\color{blue}{25}&\color{orangered}{48} \end{array} $$The solution is:
$$ \frac{ x^{4}+3x^{3}+x^{2}+3x-2 }{ x-2 } = \color{blue}{x^{3}+5x^{2}+11x+25} ~+~ \frac{ \color{red}{ 48 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&3&1&3&-2\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 1 }&3&1&3&-2\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&3&1&3&-2\\& & \color{blue}{2} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 2 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrr}2&1&\color{orangered}{ 3 }&1&3&-2\\& & \color{orangered}{2} & & & \\ \hline &1&\color{orangered}{5}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&3&1&3&-2\\& & 2& \color{blue}{10} & & \\ \hline &1&\color{blue}{5}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 10 } = \color{orangered}{ 11 } $
$$ \begin{array}{c|rrrrr}2&1&3&\color{orangered}{ 1 }&3&-2\\& & 2& \color{orangered}{10} & & \\ \hline &1&5&\color{orangered}{11}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 11 } = \color{blue}{ 22 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&3&1&3&-2\\& & 2& 10& \color{blue}{22} & \\ \hline &1&5&\color{blue}{11}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 22 } = \color{orangered}{ 25 } $
$$ \begin{array}{c|rrrrr}2&1&3&1&\color{orangered}{ 3 }&-2\\& & 2& 10& \color{orangered}{22} & \\ \hline &1&5&11&\color{orangered}{25}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 25 } = \color{blue}{ 50 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&3&1&3&-2\\& & 2& 10& 22& \color{blue}{50} \\ \hline &1&5&11&\color{blue}{25}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 50 } = \color{orangered}{ 48 } $
$$ \begin{array}{c|rrrrr}2&1&3&1&3&\color{orangered}{ -2 }\\& & 2& 10& 22& \color{orangered}{50} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{11}&\color{blue}{25}&\color{orangered}{48} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+5x^{2}+11x+25 } $ with a remainder of $ \color{red}{ 48 } $.