Divide using synthetic division $$ ( x^{4}+8x^{3}+10x^{2}+2x+4 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&1&8&10&2&4\\& & 2& 20& 60& \color{black}{124} \\ \hline &\color{blue}{1}&\color{blue}{10}&\color{blue}{30}&\color{blue}{62}&\color{orangered}{128} \end{array} $$The solution is:
$$ \frac{ x^{4}+8x^{3}+10x^{2}+2x+4 }{ x-2 } = \color{blue}{x^{3}+10x^{2}+30x+62} ~+~ \frac{ \color{red}{ 128 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&8&10&2&4\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 1 }&8&10&2&4\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 1 } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&8&10&2&4\\& & \color{blue}{2} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ 2 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrr}2&1&\color{orangered}{ 8 }&10&2&4\\& & \color{orangered}{2} & & & \\ \hline &1&\color{orangered}{10}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&8&10&2&4\\& & 2& \color{blue}{20} & & \\ \hline &1&\color{blue}{10}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ 20 } = \color{orangered}{ 30 } $
$$ \begin{array}{c|rrrrr}2&1&8&\color{orangered}{ 10 }&2&4\\& & 2& \color{orangered}{20} & & \\ \hline &1&10&\color{orangered}{30}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 30 } = \color{blue}{ 60 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&8&10&2&4\\& & 2& 20& \color{blue}{60} & \\ \hline &1&10&\color{blue}{30}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 60 } = \color{orangered}{ 62 } $
$$ \begin{array}{c|rrrrr}2&1&8&10&\color{orangered}{ 2 }&4\\& & 2& 20& \color{orangered}{60} & \\ \hline &1&10&30&\color{orangered}{62}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 62 } = \color{blue}{ 124 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&8&10&2&4\\& & 2& 20& 60& \color{blue}{124} \\ \hline &1&10&30&\color{blue}{62}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 124 } = \color{orangered}{ 128 } $
$$ \begin{array}{c|rrrrr}2&1&8&10&2&\color{orangered}{ 4 }\\& & 2& 20& 60& \color{orangered}{124} \\ \hline &\color{blue}{1}&\color{blue}{10}&\color{blue}{30}&\color{blue}{62}&\color{orangered}{128} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+10x^{2}+30x+62 } $ with a remainder of $ \color{red}{ 128 } $.