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