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