Divide using synthetic division $$ ( x^{4}+2x^{3}+3x^{2}-4x+5 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&1&2&3&-4&5\\& & 2& 8& 22& \color{black}{36} \\ \hline &\color{blue}{1}&\color{blue}{4}&\color{blue}{11}&\color{blue}{18}&\color{orangered}{41} \end{array} $$The solution is:
$$ \frac{ x^{4}+2x^{3}+3x^{2}-4x+5 }{ x-2 } = \color{blue}{x^{3}+4x^{2}+11x+18} ~+~ \frac{ \color{red}{ 41 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&2&3&-4&5\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 1 }&2&3&-4&5\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&2&3&-4&5\\& & \color{blue}{2} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 2 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}2&1&\color{orangered}{ 2 }&3&-4&5\\& & \color{orangered}{2} & & & \\ \hline &1&\color{orangered}{4}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&2&3&-4&5\\& & 2& \color{blue}{8} & & \\ \hline &1&\color{blue}{4}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 8 } = \color{orangered}{ 11 } $
$$ \begin{array}{c|rrrrr}2&1&2&\color{orangered}{ 3 }&-4&5\\& & 2& \color{orangered}{8} & & \\ \hline &1&4&\color{orangered}{11}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&2&3&-4&5\\& & 2& 8& \color{blue}{22} & \\ \hline &1&4&\color{blue}{11}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 22 } = \color{orangered}{ 18 } $
$$ \begin{array}{c|rrrrr}2&1&2&3&\color{orangered}{ -4 }&5\\& & 2& 8& \color{orangered}{22} & \\ \hline &1&4&11&\color{orangered}{18}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 18 } = \color{blue}{ 36 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&2&3&-4&5\\& & 2& 8& 22& \color{blue}{36} \\ \hline &1&4&11&\color{blue}{18}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 36 } = \color{orangered}{ 41 } $
$$ \begin{array}{c|rrrrr}2&1&2&3&-4&\color{orangered}{ 5 }\\& & 2& 8& 22& \color{orangered}{36} \\ \hline &\color{blue}{1}&\color{blue}{4}&\color{blue}{11}&\color{blue}{18}&\color{orangered}{41} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+4x^{2}+11x+18 } $ with a remainder of $ \color{red}{ 41 } $.