Divide using synthetic division $$ ( 3x^{5}-4x^{3}+5x^{2}-2x+4 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}2&3&0&-4&5&-2&4\\& & 6& 12& 16& 42& \color{black}{80} \\ \hline &\color{blue}{3}&\color{blue}{6}&\color{blue}{8}&\color{blue}{21}&\color{blue}{40}&\color{orangered}{84} \end{array} $$The solution is:
$$ \frac{ 3x^{5}-4x^{3}+5x^{2}-2x+4 }{ x-2 } = \color{blue}{3x^{4}+6x^{3}+8x^{2}+21x+40} ~+~ \frac{ \color{red}{ 84 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&3&0&-4&5&-2&4\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}2&\color{orangered}{ 3 }&0&-4&5&-2&4\\& & & & & & \\ \hline &\color{orangered}{3}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 3 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&3&0&-4&5&-2&4\\& & \color{blue}{6} & & & & \\ \hline &\color{blue}{3}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 6 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrrr}2&3&\color{orangered}{ 0 }&-4&5&-2&4\\& & \color{orangered}{6} & & & & \\ \hline &3&\color{orangered}{6}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 6 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&3&0&-4&5&-2&4\\& & 6& \color{blue}{12} & & & \\ \hline &3&\color{blue}{6}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 12 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrrr}2&3&0&\color{orangered}{ -4 }&5&-2&4\\& & 6& \color{orangered}{12} & & & \\ \hline &3&6&\color{orangered}{8}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 8 } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&3&0&-4&5&-2&4\\& & 6& 12& \color{blue}{16} & & \\ \hline &3&6&\color{blue}{8}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 16 } = \color{orangered}{ 21 } $
$$ \begin{array}{c|rrrrrr}2&3&0&-4&\color{orangered}{ 5 }&-2&4\\& & 6& 12& \color{orangered}{16} & & \\ \hline &3&6&8&\color{orangered}{21}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 21 } = \color{blue}{ 42 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&3&0&-4&5&-2&4\\& & 6& 12& 16& \color{blue}{42} & \\ \hline &3&6&8&\color{blue}{21}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 42 } = \color{orangered}{ 40 } $
$$ \begin{array}{c|rrrrrr}2&3&0&-4&5&\color{orangered}{ -2 }&4\\& & 6& 12& 16& \color{orangered}{42} & \\ \hline &3&6&8&21&\color{orangered}{40}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 40 } = \color{blue}{ 80 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&3&0&-4&5&-2&4\\& & 6& 12& 16& 42& \color{blue}{80} \\ \hline &3&6&8&21&\color{blue}{40}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 80 } = \color{orangered}{ 84 } $
$$ \begin{array}{c|rrrrrr}2&3&0&-4&5&-2&\color{orangered}{ 4 }\\& & 6& 12& 16& 42& \color{orangered}{80} \\ \hline &\color{blue}{3}&\color{blue}{6}&\color{blue}{8}&\color{blue}{21}&\color{blue}{40}&\color{orangered}{84} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{4}+6x^{3}+8x^{2}+21x+40 } $ with a remainder of $ \color{red}{ 84 } $.