Divide using synthetic division $$ ( 4x^{4}+5x^{2}+2x+4 ) \div ( 2x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}1&4&0&5&2&4\\& & 4& 4& 9& \color{black}{11} \\ \hline &\color{blue}{4}&\color{blue}{4}&\color{blue}{9}&\color{blue}{11}&\color{orangered}{15} \end{array} $$The solution is:
$$ \frac{ 4x^{4}+5x^{2}+2x+4 }{ x-1 } = \color{blue}{4x^{3}+4x^{2}+9x+11} ~+~ \frac{ \color{red}{ 15 } }{ x-1 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&4&0&5&2&4\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}1&\color{orangered}{ 4 }&0&5&2&4\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 4 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&4&0&5&2&4\\& & \color{blue}{4} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 4 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}1&4&\color{orangered}{ 0 }&5&2&4\\& & \color{orangered}{4} & & & \\ \hline &4&\color{orangered}{4}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 4 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&4&0&5&2&4\\& & 4& \color{blue}{4} & & \\ \hline &4&\color{blue}{4}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 4 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrr}1&4&0&\color{orangered}{ 5 }&2&4\\& & 4& \color{orangered}{4} & & \\ \hline &4&4&\color{orangered}{9}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 9 } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&4&0&5&2&4\\& & 4& 4& \color{blue}{9} & \\ \hline &4&4&\color{blue}{9}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 9 } = \color{orangered}{ 11 } $
$$ \begin{array}{c|rrrrr}1&4&0&5&\color{orangered}{ 2 }&4\\& & 4& 4& \color{orangered}{9} & \\ \hline &4&4&9&\color{orangered}{11}& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&4&0&5&2&4\\& & 4& 4& 9& \color{blue}{11} \\ \hline &4&4&9&\color{blue}{11}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 11 } = \color{orangered}{ 15 } $
$$ \begin{array}{c|rrrrr}1&4&0&5&2&\color{orangered}{ 4 }\\& & 4& 4& 9& \color{orangered}{11} \\ \hline &\color{blue}{4}&\color{blue}{4}&\color{blue}{9}&\color{blue}{11}&\color{orangered}{15} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{3}+4x^{2}+9x+11 } $ with a remainder of $ \color{red}{ 15 } $.