Divide using synthetic division $$ ( 2x^{4}+3x^{2}+5x+1 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&2&0&3&5&1\\& & 4& 8& 22& \color{black}{54} \\ \hline &\color{blue}{2}&\color{blue}{4}&\color{blue}{11}&\color{blue}{27}&\color{orangered}{55} \end{array} $$The solution is:
$$ \frac{ 2x^{4}+3x^{2}+5x+1 }{ x-2 } = \color{blue}{2x^{3}+4x^{2}+11x+27} ~+~ \frac{ \color{red}{ 55 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&2&0&3&5&1\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 2 }&0&3&5&1\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 2 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&2&0&3&5&1\\& & \color{blue}{4} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 4 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}2&2&\color{orangered}{ 0 }&3&5&1\\& & \color{orangered}{4} & & & \\ \hline &2&\color{orangered}{4}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&2&0&3&5&1\\& & 4& \color{blue}{8} & & \\ \hline &2&\color{blue}{4}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 8 } = \color{orangered}{ 11 } $
$$ \begin{array}{c|rrrrr}2&2&0&\color{orangered}{ 3 }&5&1\\& & 4& \color{orangered}{8} & & \\ \hline &2&4&\color{orangered}{11}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&2&0&3&5&1\\& & 4& 8& \color{blue}{22} & \\ \hline &2&4&\color{blue}{11}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 22 } = \color{orangered}{ 27 } $
$$ \begin{array}{c|rrrrr}2&2&0&3&\color{orangered}{ 5 }&1\\& & 4& 8& \color{orangered}{22} & \\ \hline &2&4&11&\color{orangered}{27}& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&2&0&3&5&1\\& & 4& 8& 22& \color{blue}{54} \\ \hline &2&4&11&\color{blue}{27}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 54 } = \color{orangered}{ 55 } $
$$ \begin{array}{c|rrrrr}2&2&0&3&5&\color{orangered}{ 1 }\\& & 4& 8& 22& \color{orangered}{54} \\ \hline &\color{blue}{2}&\color{blue}{4}&\color{blue}{11}&\color{blue}{27}&\color{orangered}{55} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}+4x^{2}+11x+27 } $ with a remainder of $ \color{red}{ 55 } $.