Divide using synthetic division $$ ( 2x^{5}+x^{2}+4x-9 ) \div ( x+1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-1&2&0&0&1&4&-9\\& & -2& 2& -2& 1& \color{black}{-5} \\ \hline &\color{blue}{2}&\color{blue}{-2}&\color{blue}{2}&\color{blue}{-1}&\color{blue}{5}&\color{orangered}{-14} \end{array} $$The solution is:
$$ \frac{ 2x^{5}+x^{2}+4x-9 }{ x+1 } = \color{blue}{2x^{4}-2x^{3}+2x^{2}-x+5} \color{red}{~-~} \frac{ \color{red}{ 14 } }{ x+1 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{-1}&2&0&0&1&4&-9\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-1&\color{orangered}{ 2 }&0&0&1&4&-9\\& & & & & & \\ \hline &\color{orangered}{2}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-1}&2&0&0&1&4&-9\\& & \color{blue}{-2} & & & & \\ \hline &\color{blue}{2}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrrr}-1&2&\color{orangered}{ 0 }&0&1&4&-9\\& & \color{orangered}{-2} & & & & \\ \hline &2&\color{orangered}{-2}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-1}&2&0&0&1&4&-9\\& & -2& \color{blue}{2} & & & \\ \hline &2&\color{blue}{-2}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 2 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrrr}-1&2&0&\color{orangered}{ 0 }&1&4&-9\\& & -2& \color{orangered}{2} & & & \\ \hline &2&-2&\color{orangered}{2}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 2 } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-1}&2&0&0&1&4&-9\\& & -2& 2& \color{blue}{-2} & & \\ \hline &2&-2&\color{blue}{2}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrrr}-1&2&0&0&\color{orangered}{ 1 }&4&-9\\& & -2& 2& \color{orangered}{-2} & & \\ \hline &2&-2&2&\color{orangered}{-1}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 1 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-1}&2&0&0&1&4&-9\\& & -2& 2& -2& \color{blue}{1} & \\ \hline &2&-2&2&\color{blue}{-1}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 1 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrrr}-1&2&0&0&1&\color{orangered}{ 4 }&-9\\& & -2& 2& -2& \color{orangered}{1} & \\ \hline &2&-2&2&-1&\color{orangered}{5}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 5 } = \color{blue}{ -5 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-1}&2&0&0&1&4&-9\\& & -2& 2& -2& 1& \color{blue}{-5} \\ \hline &2&-2&2&-1&\color{blue}{5}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -9 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ -14 } $
$$ \begin{array}{c|rrrrrr}-1&2&0&0&1&4&\color{orangered}{ -9 }\\& & -2& 2& -2& 1& \color{orangered}{-5} \\ \hline &\color{blue}{2}&\color{blue}{-2}&\color{blue}{2}&\color{blue}{-1}&\color{blue}{5}&\color{orangered}{-14} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{4}-2x^{3}+2x^{2}-x+5 } $ with a remainder of $ \color{red}{ -14 } $.