Find remainder when $ x^{3}+16x^{2}+60x+40 $ is divided by $ x+2 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&1&16&60&40\\& & -2& -28& \color{black}{-64} \\ \hline &\color{blue}{1}&\color{blue}{14}&\color{blue}{32}&\color{orangered}{-24} \end{array} $$The remainder when $ x^{3}+16x^{2}+60x+40 $ is divided by $ x+2 $ is $ \, \color{red}{ -24 } $.
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 + 2 = 0 $ ( $ x = \color{blue}{ -2 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&16&60&40\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 1 }&16&60&40\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 1 } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&16&60&40\\& & \color{blue}{-2} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ 14 } $
$$ \begin{array}{c|rrrr}-2&1&\color{orangered}{ 16 }&60&40\\& & \color{orangered}{-2} & & \\ \hline &1&\color{orangered}{14}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 14 } = \color{blue}{ -28 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&16&60&40\\& & -2& \color{blue}{-28} & \\ \hline &1&\color{blue}{14}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 60 } + \color{orangered}{ \left( -28 \right) } = \color{orangered}{ 32 } $
$$ \begin{array}{c|rrrr}-2&1&16&\color{orangered}{ 60 }&40\\& & -2& \color{orangered}{-28} & \\ \hline &1&14&\color{orangered}{32}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 32 } = \color{blue}{ -64 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&16&60&40\\& & -2& -28& \color{blue}{-64} \\ \hline &1&14&\color{blue}{32}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 40 } + \color{orangered}{ \left( -64 \right) } = \color{orangered}{ -24 } $
$$ \begin{array}{c|rrrr}-2&1&16&60&\color{orangered}{ 40 }\\& & -2& -28& \color{orangered}{-64} \\ \hline &\color{blue}{1}&\color{blue}{14}&\color{blue}{32}&\color{orangered}{-24} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -24 }\right) $.