Find remainder when $ x^{3}+x^{2}-16x-16 $ is divided by $ x-4 $.

The synthetic division table is:

$$ \begin{array}{c|rrrr}4&1&1&-16&-16\\& & 4& 20& \color{black}{16} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{4}&\color{orangered}{0} \end{array} $$

The remainder when $ x^{3}+x^{2}-16x-16 $ is divided by $ x-4 $ is $ \, \color{red}{ 0 } $.

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 -4 = 0 $ ( $ x = \color{blue}{ 4 } $ ) at the left.

$$ \begin{array}{c|rrrr}\color{blue}{4}&1&1&-16&-16\\& & & & \\ \hline &&&& \end{array} $$

Step 1 : Bring down the leading coefficient to the bottom row.

$$ \begin{array}{c|rrrr}4&\color{orangered}{ 1 }&1&-16&-16\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$

Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 1 } = \color{blue}{ 4 } $.

$$ \begin{array}{c|rrrr}\color{blue}{4}&1&1&-16&-16\\& & \color{blue}{4} & & \\ \hline &\color{blue}{1}&&& \end{array} $$

Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 4 } = \color{orangered}{ 5 } $

$$ \begin{array}{c|rrrr}4&1&\color{orangered}{ 1 }&-16&-16\\& & \color{orangered}{4} & & \\ \hline &1&\color{orangered}{5}&& \end{array} $$

Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 5 } = \color{blue}{ 20 } $.

$$ \begin{array}{c|rrrr}\color{blue}{4}&1&1&-16&-16\\& & 4& \color{blue}{20} & \\ \hline &1&\color{blue}{5}&& \end{array} $$

Step 5 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 20 } = \color{orangered}{ 4 } $

$$ \begin{array}{c|rrrr}4&1&1&\color{orangered}{ -16 }&-16\\& & 4& \color{orangered}{20} & \\ \hline &1&5&\color{orangered}{4}& \end{array} $$

Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 4 } = \color{blue}{ 16 } $.

$$ \begin{array}{c|rrrr}\color{blue}{4}&1&1&-16&-16\\& & 4& 20& \color{blue}{16} \\ \hline &1&5&\color{blue}{4}& \end{array} $$

Step 7 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 16 } = \color{orangered}{ 0 } $

$$ \begin{array}{c|rrrr}4&1&1&-16&\color{orangered}{ -16 }\\& & 4& 20& \color{orangered}{16} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{4}&\color{orangered}{0} \end{array} $$

Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right) $.

Synthetic Division Calculator