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