Find remainder when $ x^{3}-16x^{2}+18x+115 $ is divided by $ x-3 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&1&-16&18&115\\& & 3& -39& \color{black}{-63} \\ \hline &\color{blue}{1}&\color{blue}{-13}&\color{blue}{-21}&\color{orangered}{52} \end{array} $$The remainder when $ x^{3}-16x^{2}+18x+115 $ is divided by $ x-3 $ is $ \, \color{red}{ 52 } $.
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 -3 = 0 $ ( $ x = \color{blue}{ 3 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-16&18&115\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 1 }&-16&18&115\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 1 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-16&18&115\\& & \color{blue}{3} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 3 } = \color{orangered}{ -13 } $
$$ \begin{array}{c|rrrr}3&1&\color{orangered}{ -16 }&18&115\\& & \color{orangered}{3} & & \\ \hline &1&\color{orangered}{-13}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -13 \right) } = \color{blue}{ -39 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-16&18&115\\& & 3& \color{blue}{-39} & \\ \hline &1&\color{blue}{-13}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 18 } + \color{orangered}{ \left( -39 \right) } = \color{orangered}{ -21 } $
$$ \begin{array}{c|rrrr}3&1&-16&\color{orangered}{ 18 }&115\\& & 3& \color{orangered}{-39} & \\ \hline &1&-13&\color{orangered}{-21}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -21 \right) } = \color{blue}{ -63 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-16&18&115\\& & 3& -39& \color{blue}{-63} \\ \hline &1&-13&\color{blue}{-21}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 115 } + \color{orangered}{ \left( -63 \right) } = \color{orangered}{ 52 } $
$$ \begin{array}{c|rrrr}3&1&-16&18&\color{orangered}{ 115 }\\& & 3& -39& \color{orangered}{-63} \\ \hline &\color{blue}{1}&\color{blue}{-13}&\color{blue}{-21}&\color{orangered}{52} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 52 }\right) $.