Find remainder when $ 3x^{3}+7x^{2}-116x+160 $ is divided by $ x-3 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&3&7&-116&160\\& & 9& 48& \color{black}{-204} \\ \hline &\color{blue}{3}&\color{blue}{16}&\color{blue}{-68}&\color{orangered}{-44} \end{array} $$The remainder when $ 3x^{3}+7x^{2}-116x+160 $ is divided by $ x-3 $ is $ \, \color{red}{ -44 } $.
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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&3&7&-116&160\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 3 }&7&-116&160\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 3 } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&3&7&-116&160\\& & \color{blue}{9} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 9 } = \color{orangered}{ 16 } $
$$ \begin{array}{c|rrrr}3&3&\color{orangered}{ 7 }&-116&160\\& & \color{orangered}{9} & & \\ \hline &3&\color{orangered}{16}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 16 } = \color{blue}{ 48 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&3&7&-116&160\\& & 9& \color{blue}{48} & \\ \hline &3&\color{blue}{16}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -116 } + \color{orangered}{ 48 } = \color{orangered}{ -68 } $
$$ \begin{array}{c|rrrr}3&3&7&\color{orangered}{ -116 }&160\\& & 9& \color{orangered}{48} & \\ \hline &3&16&\color{orangered}{-68}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -68 \right) } = \color{blue}{ -204 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&3&7&-116&160\\& & 9& 48& \color{blue}{-204} \\ \hline &3&16&\color{blue}{-68}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 160 } + \color{orangered}{ \left( -204 \right) } = \color{orangered}{ -44 } $
$$ \begin{array}{c|rrrr}3&3&7&-116&\color{orangered}{ 160 }\\& & 9& 48& \color{orangered}{-204} \\ \hline &\color{blue}{3}&\color{blue}{16}&\color{blue}{-68}&\color{orangered}{-44} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -44 }\right) $.