Divide using synthetic division $$ ( x^{4}-5x^{3}+6x^{2}-4 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&1&-5&6&0&-4\\& & -2& 14& -40& \color{black}{80} \\ \hline &\color{blue}{1}&\color{blue}{-7}&\color{blue}{20}&\color{blue}{-40}&\color{orangered}{76} \end{array} $$The solution is:
$$ \frac{ x^{4}-5x^{3}+6x^{2}-4 }{ x+2 } = \color{blue}{x^{3}-7x^{2}+20x-40} ~+~ \frac{ \color{red}{ 76 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&-5&6&0&-4\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 1 }&-5&6&0&-4\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&-5&6&0&-4\\& & \color{blue}{-2} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrrr}-2&1&\color{orangered}{ -5 }&6&0&-4\\& & \color{orangered}{-2} & & & \\ \hline &1&\color{orangered}{-7}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ 14 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&-5&6&0&-4\\& & -2& \color{blue}{14} & & \\ \hline &1&\color{blue}{-7}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 14 } = \color{orangered}{ 20 } $
$$ \begin{array}{c|rrrrr}-2&1&-5&\color{orangered}{ 6 }&0&-4\\& & -2& \color{orangered}{14} & & \\ \hline &1&-7&\color{orangered}{20}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 20 } = \color{blue}{ -40 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&-5&6&0&-4\\& & -2& 14& \color{blue}{-40} & \\ \hline &1&-7&\color{blue}{20}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -40 \right) } = \color{orangered}{ -40 } $
$$ \begin{array}{c|rrrrr}-2&1&-5&6&\color{orangered}{ 0 }&-4\\& & -2& 14& \color{orangered}{-40} & \\ \hline &1&-7&20&\color{orangered}{-40}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -40 \right) } = \color{blue}{ 80 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&-5&6&0&-4\\& & -2& 14& -40& \color{blue}{80} \\ \hline &1&-7&20&\color{blue}{-40}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 80 } = \color{orangered}{ 76 } $
$$ \begin{array}{c|rrrrr}-2&1&-5&6&0&\color{orangered}{ -4 }\\& & -2& 14& -40& \color{orangered}{80} \\ \hline &\color{blue}{1}&\color{blue}{-7}&\color{blue}{20}&\color{blue}{-40}&\color{orangered}{76} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-7x^{2}+20x-40 } $ with a remainder of $ \color{red}{ 76 } $.