Divide using synthetic division $$ ( 4x^{4}-3x^{3}-x^{2}+2x+1 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&4&-3&-1&2&1\\& & 8& 10& 18& \color{black}{40} \\ \hline &\color{blue}{4}&\color{blue}{5}&\color{blue}{9}&\color{blue}{20}&\color{orangered}{41} \end{array} $$The solution is:
$$ \frac{ 4x^{4}-3x^{3}-x^{2}+2x+1 }{ x-2 } = \color{blue}{4x^{3}+5x^{2}+9x+20} ~+~ \frac{ \color{red}{ 41 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&4&-3&-1&2&1\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 4 }&-3&-1&2&1\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 4 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&4&-3&-1&2&1\\& & \color{blue}{8} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 8 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrr}2&4&\color{orangered}{ -3 }&-1&2&1\\& & \color{orangered}{8} & & & \\ \hline &4&\color{orangered}{5}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 5 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&4&-3&-1&2&1\\& & 8& \color{blue}{10} & & \\ \hline &4&\color{blue}{5}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 10 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrr}2&4&-3&\color{orangered}{ -1 }&2&1\\& & 8& \color{orangered}{10} & & \\ \hline &4&5&\color{orangered}{9}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 9 } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&4&-3&-1&2&1\\& & 8& 10& \color{blue}{18} & \\ \hline &4&5&\color{blue}{9}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 18 } = \color{orangered}{ 20 } $
$$ \begin{array}{c|rrrrr}2&4&-3&-1&\color{orangered}{ 2 }&1\\& & 8& 10& \color{orangered}{18} & \\ \hline &4&5&9&\color{orangered}{20}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 20 } = \color{blue}{ 40 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&4&-3&-1&2&1\\& & 8& 10& 18& \color{blue}{40} \\ \hline &4&5&9&\color{blue}{20}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 40 } = \color{orangered}{ 41 } $
$$ \begin{array}{c|rrrrr}2&4&-3&-1&2&\color{orangered}{ 1 }\\& & 8& 10& 18& \color{orangered}{40} \\ \hline &\color{blue}{4}&\color{blue}{5}&\color{blue}{9}&\color{blue}{20}&\color{orangered}{41} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{3}+5x^{2}+9x+20 } $ with a remainder of $ \color{red}{ 41 } $.