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