Find remainder when $ 2x^{3}-x^{2}-x+2 $ is divided by $ x+1 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-1&2&-1&-1&2\\& & -2& 3& \color{black}{-2} \\ \hline &\color{blue}{2}&\color{blue}{-3}&\color{blue}{2}&\color{orangered}{0} \end{array} $$The remainder when $ 2x^{3}-x^{2}-x+2 $ is divided by $ x+1 $ is $ \, \color{red}{ 0 } $.
Explanation
We can find remainder using synthetic division method.
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.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&2&-1&-1&2\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-1&\color{orangered}{ 2 }&-1&-1&2\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&2&-1&-1&2\\& & \color{blue}{-2} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrr}-1&2&\color{orangered}{ -1 }&-1&2\\& & \color{orangered}{-2} & & \\ \hline &2&\color{orangered}{-3}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&2&-1&-1&2\\& & -2& \color{blue}{3} & \\ \hline &2&\color{blue}{-3}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 3 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrr}-1&2&-1&\color{orangered}{ -1 }&2\\& & -2& \color{orangered}{3} & \\ \hline &2&-3&\color{orangered}{2}& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&2&-1&-1&2\\& & -2& 3& \color{blue}{-2} \\ \hline &2&-3&\color{blue}{2}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}-1&2&-1&-1&\color{orangered}{ 2 }\\& & -2& 3& \color{orangered}{-2} \\ \hline &\color{blue}{2}&\color{blue}{-3}&\color{blue}{2}&\color{orangered}{0} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right) $.