Find remainder when $ 11x^{3}-16x^{2}-11x+16 $ is divided by $ x+1 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-1&11&-16&-11&16\\& & -11& 27& \color{black}{-16} \\ \hline &\color{blue}{11}&\color{blue}{-27}&\color{blue}{16}&\color{orangered}{0} \end{array} $$The remainder when $ 11x^{3}-16x^{2}-11x+16 $ 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.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&11&-16&-11&16\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-1&\color{orangered}{ 11 }&-16&-11&16\\& & & & \\ \hline &\color{orangered}{11}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 11 } = \color{blue}{ -11 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&11&-16&-11&16\\& & \color{blue}{-11} & & \\ \hline &\color{blue}{11}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ \left( -11 \right) } = \color{orangered}{ -27 } $
$$ \begin{array}{c|rrrr}-1&11&\color{orangered}{ -16 }&-11&16\\& & \color{orangered}{-11} & & \\ \hline &11&\color{orangered}{-27}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -27 \right) } = \color{blue}{ 27 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&11&-16&-11&16\\& & -11& \color{blue}{27} & \\ \hline &11&\color{blue}{-27}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -11 } + \color{orangered}{ 27 } = \color{orangered}{ 16 } $
$$ \begin{array}{c|rrrr}-1&11&-16&\color{orangered}{ -11 }&16\\& & -11& \color{orangered}{27} & \\ \hline &11&-27&\color{orangered}{16}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 16 } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&11&-16&-11&16\\& & -11& 27& \color{blue}{-16} \\ \hline &11&-27&\color{blue}{16}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}-1&11&-16&-11&\color{orangered}{ 16 }\\& & -11& 27& \color{orangered}{-16} \\ \hline &\color{blue}{11}&\color{blue}{-27}&\color{blue}{16}&\color{orangered}{0} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right) $.