Divide using synthetic division $$ ( x^{4}+2x^{3}+5x^{2}+x+4 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&1&2&5&1&4\\& & 2& 8& 26& \color{black}{54} \\ \hline &\color{blue}{1}&\color{blue}{4}&\color{blue}{13}&\color{blue}{27}&\color{orangered}{58} \end{array} $$The solution is:
$$ \frac{ x^{4}+2x^{3}+5x^{2}+x+4 }{ x-2 } = \color{blue}{x^{3}+4x^{2}+13x+27} ~+~ \frac{ \color{red}{ 58 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&2&5&1&4\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 1 }&2&5&1&4\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&2&5&1&4\\& & \color{blue}{2} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 2 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}2&1&\color{orangered}{ 2 }&5&1&4\\& & \color{orangered}{2} & & & \\ \hline &1&\color{orangered}{4}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 4 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&2&5&1&4\\& & 2& \color{blue}{8} & & \\ \hline &1&\color{blue}{4}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 8 } = \color{orangered}{ 13 } $
$$ \begin{array}{c|rrrrr}2&1&2&\color{orangered}{ 5 }&1&4\\& & 2& \color{orangered}{8} & & \\ \hline &1&4&\color{orangered}{13}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 13 } = \color{blue}{ 26 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&2&5&1&4\\& & 2& 8& \color{blue}{26} & \\ \hline &1&4&\color{blue}{13}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 26 } = \color{orangered}{ 27 } $
$$ \begin{array}{c|rrrrr}2&1&2&5&\color{orangered}{ 1 }&4\\& & 2& 8& \color{orangered}{26} & \\ \hline &1&4&13&\color{orangered}{27}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 27 } = \color{blue}{ 54 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&2&5&1&4\\& & 2& 8& 26& \color{blue}{54} \\ \hline &1&4&13&\color{blue}{27}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 54 } = \color{orangered}{ 58 } $
$$ \begin{array}{c|rrrrr}2&1&2&5&1&\color{orangered}{ 4 }\\& & 2& 8& 26& \color{orangered}{54} \\ \hline &\color{blue}{1}&\color{blue}{4}&\color{blue}{13}&\color{blue}{27}&\color{orangered}{58} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+4x^{2}+13x+27 } $ with a remainder of $ \color{red}{ 58 } $.