Divide using synthetic division $$ ( 5x^{6}-9x^{4}+7x^{2}+4 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrrr}2&5&0&-9&0&7&0&4\\& & 10& 20& 22& 44& 102& \color{black}{204} \\ \hline &\color{blue}{5}&\color{blue}{10}&\color{blue}{11}&\color{blue}{22}&\color{blue}{51}&\color{blue}{102}&\color{orangered}{208} \end{array} $$The solution is:
$$ \frac{ 5x^{6}-9x^{4}+7x^{2}+4 }{ x-2 } = \color{blue}{5x^{5}+10x^{4}+11x^{3}+22x^{2}+51x+102} ~+~ \frac{ \color{red}{ 208 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrrrrr}\color{blue}{2}&5&0&-9&0&7&0&4\\& & & & & & & \\ \hline &&&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrrr}2&\color{orangered}{ 5 }&0&-9&0&7&0&4\\& & & & & & & \\ \hline &\color{orangered}{5}&&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 5 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{2}&5&0&-9&0&7&0&4\\& & \color{blue}{10} & & & & & \\ \hline &\color{blue}{5}&&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 10 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrrrr}2&5&\color{orangered}{ 0 }&-9&0&7&0&4\\& & \color{orangered}{10} & & & & & \\ \hline &5&\color{orangered}{10}&&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 10 } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{2}&5&0&-9&0&7&0&4\\& & 10& \color{blue}{20} & & & & \\ \hline &5&\color{blue}{10}&&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -9 } + \color{orangered}{ 20 } = \color{orangered}{ 11 } $
$$ \begin{array}{c|rrrrrrr}2&5&0&\color{orangered}{ -9 }&0&7&0&4\\& & 10& \color{orangered}{20} & & & & \\ \hline &5&10&\color{orangered}{11}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{2}&5&0&-9&0&7&0&4\\& & 10& 20& \color{blue}{22} & & & \\ \hline &5&10&\color{blue}{11}&&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 22 } = \color{orangered}{ 22 } $
$$ \begin{array}{c|rrrrrrr}2&5&0&-9&\color{orangered}{ 0 }&7&0&4\\& & 10& 20& \color{orangered}{22} & & & \\ \hline &5&10&11&\color{orangered}{22}&&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 22 } = \color{blue}{ 44 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{2}&5&0&-9&0&7&0&4\\& & 10& 20& 22& \color{blue}{44} & & \\ \hline &5&10&11&\color{blue}{22}&&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 44 } = \color{orangered}{ 51 } $
$$ \begin{array}{c|rrrrrrr}2&5&0&-9&0&\color{orangered}{ 7 }&0&4\\& & 10& 20& 22& \color{orangered}{44} & & \\ \hline &5&10&11&22&\color{orangered}{51}&& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 51 } = \color{blue}{ 102 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{2}&5&0&-9&0&7&0&4\\& & 10& 20& 22& 44& \color{blue}{102} & \\ \hline &5&10&11&22&\color{blue}{51}&& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 102 } = \color{orangered}{ 102 } $
$$ \begin{array}{c|rrrrrrr}2&5&0&-9&0&7&\color{orangered}{ 0 }&4\\& & 10& 20& 22& 44& \color{orangered}{102} & \\ \hline &5&10&11&22&51&\color{orangered}{102}& \end{array} $$Step 12 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 102 } = \color{blue}{ 204 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{2}&5&0&-9&0&7&0&4\\& & 10& 20& 22& 44& 102& \color{blue}{204} \\ \hline &5&10&11&22&51&\color{blue}{102}& \end{array} $$Step 13 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 204 } = \color{orangered}{ 208 } $
$$ \begin{array}{c|rrrrrrr}2&5&0&-9&0&7&0&\color{orangered}{ 4 }\\& & 10& 20& 22& 44& 102& \color{orangered}{204} \\ \hline &\color{blue}{5}&\color{blue}{10}&\color{blue}{11}&\color{blue}{22}&\color{blue}{51}&\color{blue}{102}&\color{orangered}{208} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{5}+10x^{4}+11x^{3}+22x^{2}+51x+102 } $ with a remainder of $ \color{red}{ 208 } $.