Find remainder when $ x^{3}-4x^{2}+5x-1 $ is divided by $ 2x+1 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-1&1&-4&5&-1\\& & -1& 5& \color{black}{-10} \\ \hline &\color{blue}{1}&\color{blue}{-5}&\color{blue}{10}&\color{orangered}{-11} \end{array} $$The remainder when $ x^{3}-4x^{2}+5x-1 $ is divided by $ x+1 $ is $ \, \color{red}{ -11 } $.
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.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&1&-4&5&-1\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-1&\color{orangered}{ 1 }&-4&5&-1\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 1 } = \color{blue}{ -1 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&1&-4&5&-1\\& & \color{blue}{-1} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ \left( -1 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}-1&1&\color{orangered}{ -4 }&5&-1\\& & \color{orangered}{-1} & & \\ \hline &1&\color{orangered}{-5}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&1&-4&5&-1\\& & -1& \color{blue}{5} & \\ \hline &1&\color{blue}{-5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 5 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrr}-1&1&-4&\color{orangered}{ 5 }&-1\\& & -1& \color{orangered}{5} & \\ \hline &1&-5&\color{orangered}{10}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 10 } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&1&-4&5&-1\\& & -1& 5& \color{blue}{-10} \\ \hline &1&-5&\color{blue}{10}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ -11 } $
$$ \begin{array}{c|rrrr}-1&1&-4&5&\color{orangered}{ -1 }\\& & -1& 5& \color{orangered}{-10} \\ \hline &\color{blue}{1}&\color{blue}{-5}&\color{blue}{10}&\color{orangered}{-11} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -11 }\right) $.