Find remainder when $ x^{3}-19x+9 $ is divided by $ x-5 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}5&1&0&-19&9\\& & 5& 25& \color{black}{30} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{6}&\color{orangered}{39} \end{array} $$The remainder when $ x^{3}-19x+9 $ is divided by $ x-5 $ is $ \, \color{red}{ 39 } $.
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 -5 = 0 $ ( $ x = \color{blue}{ 5 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{5}&1&0&-19&9\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}5&\color{orangered}{ 1 }&0&-19&9\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 1 } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&1&0&-19&9\\& & \color{blue}{5} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 5 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrr}5&1&\color{orangered}{ 0 }&-19&9\\& & \color{orangered}{5} & & \\ \hline &1&\color{orangered}{5}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 5 } = \color{blue}{ 25 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&1&0&-19&9\\& & 5& \color{blue}{25} & \\ \hline &1&\color{blue}{5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -19 } + \color{orangered}{ 25 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrr}5&1&0&\color{orangered}{ -19 }&9\\& & 5& \color{orangered}{25} & \\ \hline &1&5&\color{orangered}{6}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 6 } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&1&0&-19&9\\& & 5& 25& \color{blue}{30} \\ \hline &1&5&\color{blue}{6}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ 30 } = \color{orangered}{ 39 } $
$$ \begin{array}{c|rrrr}5&1&0&-19&\color{orangered}{ 9 }\\& & 5& 25& \color{orangered}{30} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{6}&\color{orangered}{39} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 39 }\right) $.