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