Divide using synthetic division $$ ( x^{4}+10x^{3}+21x^{2}+6x-8 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&1&10&21&6&-8\\& & 2& 24& 90& \color{black}{192} \\ \hline &\color{blue}{1}&\color{blue}{12}&\color{blue}{45}&\color{blue}{96}&\color{orangered}{184} \end{array} $$The solution is:
$$ \frac{ x^{4}+10x^{3}+21x^{2}+6x-8 }{ x-2 } = \color{blue}{x^{3}+12x^{2}+45x+96} ~+~ \frac{ \color{red}{ 184 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&10&21&6&-8\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 1 }&10&21&6&-8\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&10&21&6&-8\\& & \color{blue}{2} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ 2 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrrr}2&1&\color{orangered}{ 10 }&21&6&-8\\& & \color{orangered}{2} & & & \\ \hline &1&\color{orangered}{12}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 12 } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&10&21&6&-8\\& & 2& \color{blue}{24} & & \\ \hline &1&\color{blue}{12}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 21 } + \color{orangered}{ 24 } = \color{orangered}{ 45 } $
$$ \begin{array}{c|rrrrr}2&1&10&\color{orangered}{ 21 }&6&-8\\& & 2& \color{orangered}{24} & & \\ \hline &1&12&\color{orangered}{45}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 45 } = \color{blue}{ 90 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&10&21&6&-8\\& & 2& 24& \color{blue}{90} & \\ \hline &1&12&\color{blue}{45}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 90 } = \color{orangered}{ 96 } $
$$ \begin{array}{c|rrrrr}2&1&10&21&\color{orangered}{ 6 }&-8\\& & 2& 24& \color{orangered}{90} & \\ \hline &1&12&45&\color{orangered}{96}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 96 } = \color{blue}{ 192 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&10&21&6&-8\\& & 2& 24& 90& \color{blue}{192} \\ \hline &1&12&45&\color{blue}{96}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 192 } = \color{orangered}{ 184 } $
$$ \begin{array}{c|rrrrr}2&1&10&21&6&\color{orangered}{ -8 }\\& & 2& 24& 90& \color{orangered}{192} \\ \hline &\color{blue}{1}&\color{blue}{12}&\color{blue}{45}&\color{blue}{96}&\color{orangered}{184} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+12x^{2}+45x+96 } $ with a remainder of $ \color{red}{ 184 } $.