Find remainder when $ x^{3}+x^{2}-16x-16 $ is divided by $ x-16 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}16&1&1&-16&-16\\& & 16& 272& \color{black}{4096} \\ \hline &\color{blue}{1}&\color{blue}{17}&\color{blue}{256}&\color{orangered}{4080} \end{array} $$The remainder when $ x^{3}+x^{2}-16x-16 $ is divided by $ x-16 $ is $ \, \color{red}{ 4080 } $.
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 -16 = 0 $ ( $ x = \color{blue}{ 16 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{16}&1&1&-16&-16\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}16&\color{orangered}{ 1 }&1&-16&-16\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 16 } \cdot \color{blue}{ 1 } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrr}\color{blue}{16}&1&1&-16&-16\\& & \color{blue}{16} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 16 } = \color{orangered}{ 17 } $
$$ \begin{array}{c|rrrr}16&1&\color{orangered}{ 1 }&-16&-16\\& & \color{orangered}{16} & & \\ \hline &1&\color{orangered}{17}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 16 } \cdot \color{blue}{ 17 } = \color{blue}{ 272 } $.
$$ \begin{array}{c|rrrr}\color{blue}{16}&1&1&-16&-16\\& & 16& \color{blue}{272} & \\ \hline &1&\color{blue}{17}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 272 } = \color{orangered}{ 256 } $
$$ \begin{array}{c|rrrr}16&1&1&\color{orangered}{ -16 }&-16\\& & 16& \color{orangered}{272} & \\ \hline &1&17&\color{orangered}{256}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 16 } \cdot \color{blue}{ 256 } = \color{blue}{ 4096 } $.
$$ \begin{array}{c|rrrr}\color{blue}{16}&1&1&-16&-16\\& & 16& 272& \color{blue}{4096} \\ \hline &1&17&\color{blue}{256}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 4096 } = \color{orangered}{ 4080 } $
$$ \begin{array}{c|rrrr}16&1&1&-16&\color{orangered}{ -16 }\\& & 16& 272& \color{orangered}{4096} \\ \hline &\color{blue}{1}&\color{blue}{17}&\color{blue}{256}&\color{orangered}{4080} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 4080 }\right) $.