Divide using synthetic division $$ ( x^{4}-2x^{3}-5x^{2}+6 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&1&-2&-5&0&6\\& & 2& 0& -10& \color{black}{-20} \\ \hline &\color{blue}{1}&\color{blue}{0}&\color{blue}{-5}&\color{blue}{-10}&\color{orangered}{-14} \end{array} $$The solution is:
$$ \frac{ x^{4}-2x^{3}-5x^{2}+6 }{ x-2 } = \color{blue}{x^{3}-5x-10} \color{red}{~-~} \frac{ \color{red}{ 14 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&-2&-5&0&6\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 1 }&-2&-5&0&6\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&-2&-5&0&6\\& & \color{blue}{2} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 2 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}2&1&\color{orangered}{ -2 }&-5&0&6\\& & \color{orangered}{2} & & & \\ \hline &1&\color{orangered}{0}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&-2&-5&0&6\\& & 2& \color{blue}{0} & & \\ \hline &1&\color{blue}{0}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 0 } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrr}2&1&-2&\color{orangered}{ -5 }&0&6\\& & 2& \color{orangered}{0} & & \\ \hline &1&0&\color{orangered}{-5}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&-2&-5&0&6\\& & 2& 0& \color{blue}{-10} & \\ \hline &1&0&\color{blue}{-5}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrrr}2&1&-2&-5&\color{orangered}{ 0 }&6\\& & 2& 0& \color{orangered}{-10} & \\ \hline &1&0&-5&\color{orangered}{-10}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&-2&-5&0&6\\& & 2& 0& -10& \color{blue}{-20} \\ \hline &1&0&-5&\color{blue}{-10}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ -14 } $
$$ \begin{array}{c|rrrrr}2&1&-2&-5&0&\color{orangered}{ 6 }\\& & 2& 0& -10& \color{orangered}{-20} \\ \hline &\color{blue}{1}&\color{blue}{0}&\color{blue}{-5}&\color{blue}{-10}&\color{orangered}{-14} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-5x-10 } $ with a remainder of $ \color{red}{ -14 } $.