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