Find remainder when $ x^{3}-2x^{2}-9x+6 $ is divided by $ x-4 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}4&1&-2&-9&6\\& & 4& 8& \color{black}{-4} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{-1}&\color{orangered}{2} \end{array} $$The remainder when $ x^{3}-2x^{2}-9x+6 $ is divided by $ x-4 $ is $ \, \color{red}{ 2 } $.
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 -4 = 0 $ ( $ x = \color{blue}{ 4 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{4}&1&-2&-9&6\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}4&\color{orangered}{ 1 }&-2&-9&6\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&1&-2&-9&6\\& & \color{blue}{4} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 4 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrr}4&1&\color{orangered}{ -2 }&-9&6\\& & \color{orangered}{4} & & \\ \hline &1&\color{orangered}{2}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 2 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&1&-2&-9&6\\& & 4& \color{blue}{8} & \\ \hline &1&\color{blue}{2}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -9 } + \color{orangered}{ 8 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrr}4&1&-2&\color{orangered}{ -9 }&6\\& & 4& \color{orangered}{8} & \\ \hline &1&2&\color{orangered}{-1}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&1&-2&-9&6\\& & 4& 8& \color{blue}{-4} \\ \hline &1&2&\color{blue}{-1}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrr}4&1&-2&-9&\color{orangered}{ 6 }\\& & 4& 8& \color{orangered}{-4} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{-1}&\color{orangered}{2} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 2 }\right) $.