Divide using synthetic division $$ ( x^{4}-2x^{3}+x+1 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&1&-2&0&1&1\\& & -2& 8& -16& \color{black}{30} \\ \hline &\color{blue}{1}&\color{blue}{-4}&\color{blue}{8}&\color{blue}{-15}&\color{orangered}{31} \end{array} $$The solution is:
$$ \frac{ x^{4}-2x^{3}+x+1 }{ x+2 } = \color{blue}{x^{3}-4x^{2}+8x-15} ~+~ \frac{ \color{red}{ 31 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&-2&0&1&1\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 1 }&-2&0&1&1\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&-2&0&1&1\\& & \color{blue}{-2} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrr}-2&1&\color{orangered}{ -2 }&0&1&1\\& & \color{orangered}{-2} & & & \\ \hline &1&\color{orangered}{-4}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&-2&0&1&1\\& & -2& \color{blue}{8} & & \\ \hline &1&\color{blue}{-4}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 8 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrr}-2&1&-2&\color{orangered}{ 0 }&1&1\\& & -2& \color{orangered}{8} & & \\ \hline &1&-4&\color{orangered}{8}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 8 } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&-2&0&1&1\\& & -2& 8& \color{blue}{-16} & \\ \hline &1&-4&\color{blue}{8}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ -15 } $
$$ \begin{array}{c|rrrrr}-2&1&-2&0&\color{orangered}{ 1 }&1\\& & -2& 8& \color{orangered}{-16} & \\ \hline &1&-4&8&\color{orangered}{-15}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -15 \right) } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&-2&0&1&1\\& & -2& 8& -16& \color{blue}{30} \\ \hline &1&-4&8&\color{blue}{-15}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 30 } = \color{orangered}{ 31 } $
$$ \begin{array}{c|rrrrr}-2&1&-2&0&1&\color{orangered}{ 1 }\\& & -2& 8& -16& \color{orangered}{30} \\ \hline &\color{blue}{1}&\color{blue}{-4}&\color{blue}{8}&\color{blue}{-15}&\color{orangered}{31} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-4x^{2}+8x-15 } $ with a remainder of $ \color{red}{ 31 } $.