Find remainder when $ x^{3}+16x^{2}+60x+40 $ is divided by $ x+1 $.
The synthetic division table is:
$$ \begin{array}{c|rrrr}-1&1&16&60&40\\& & -1& -15& \color{black}{-45} \\ \hline &\color{blue}{1}&\color{blue}{15}&\color{blue}{45}&\color{orangered}{-5} \end{array} $$The remainder when $ x^{3}+16x^{2}+60x+40 $ is divided by $ x+1 $ is $ \, \color{red}{ -5 } $.
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.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&1&16&60&40\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-1&\color{orangered}{ 1 }&16&60&40\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&1&16&60&40\\& & \color{blue}{-1} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ \left( -1 \right) } = \color{orangered}{ 15 } $
$$ \begin{array}{c|rrrr}-1&1&\color{orangered}{ 16 }&60&40\\& & \color{orangered}{-1} & & \\ \hline &1&\color{orangered}{15}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 15 } = \color{blue}{ -15 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&1&16&60&40\\& & -1& \color{blue}{-15} & \\ \hline &1&\color{blue}{15}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 60 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ 45 } $
$$ \begin{array}{c|rrrr}-1&1&16&\color{orangered}{ 60 }&40\\& & -1& \color{orangered}{-15} & \\ \hline &1&15&\color{orangered}{45}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 45 } = \color{blue}{ -45 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&1&16&60&40\\& & -1& -15& \color{blue}{-45} \\ \hline &1&15&\color{blue}{45}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 40 } + \color{orangered}{ \left( -45 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}-1&1&16&60&\color{orangered}{ 40 }\\& & -1& -15& \color{orangered}{-45} \\ \hline &\color{blue}{1}&\color{blue}{15}&\color{blue}{45}&\color{orangered}{-5} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -5 }\right) $.