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