Divide using synthetic division $$ ( x^{4}+x^{3}-10x^{2}+20 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&1&1&-10&0&20\\& & 2& 6& -8& \color{black}{-16} \\ \hline &\color{blue}{1}&\color{blue}{3}&\color{blue}{-4}&\color{blue}{-8}&\color{orangered}{4} \end{array} $$The solution is:
$$ \frac{ x^{4}+x^{3}-10x^{2}+20 }{ x-2 } = \color{blue}{x^{3}+3x^{2}-4x-8} ~+~ \frac{ \color{red}{ 4 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&1&-10&0&20\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 1 }&1&-10&0&20\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&1&-10&0&20\\& & \color{blue}{2} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 2 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}2&1&\color{orangered}{ 1 }&-10&0&20\\& & \color{orangered}{2} & & & \\ \hline &1&\color{orangered}{3}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 3 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&1&-10&0&20\\& & 2& \color{blue}{6} & & \\ \hline &1&\color{blue}{3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 6 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrr}2&1&1&\color{orangered}{ -10 }&0&20\\& & 2& \color{orangered}{6} & & \\ \hline &1&3&\color{orangered}{-4}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&1&-10&0&20\\& & 2& 6& \color{blue}{-8} & \\ \hline &1&3&\color{blue}{-4}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrr}2&1&1&-10&\color{orangered}{ 0 }&20\\& & 2& 6& \color{orangered}{-8} & \\ \hline &1&3&-4&\color{orangered}{-8}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&1&-10&0&20\\& & 2& 6& -8& \color{blue}{-16} \\ \hline &1&3&-4&\color{blue}{-8}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 20 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}2&1&1&-10&0&\color{orangered}{ 20 }\\& & 2& 6& -8& \color{orangered}{-16} \\ \hline &\color{blue}{1}&\color{blue}{3}&\color{blue}{-4}&\color{blue}{-8}&\color{orangered}{4} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+3x^{2}-4x-8 } $ with a remainder of $ \color{red}{ 4 } $.