Find remainder when $ 3x^{3}-7x^{2}-16x+7 $ is divided by $ x-4 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}4&3&-7&-16&7\\& & 12& 20& \color{black}{16} \\ \hline &\color{blue}{3}&\color{blue}{5}&\color{blue}{4}&\color{orangered}{23} \end{array} $$The remainder when $ 3x^{3}-7x^{2}-16x+7 $ is divided by $ x-4 $ is $ \, \color{red}{ 23 } $.
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.
$$ \begin{array}{c|rrrr}\color{blue}{4}&3&-7&-16&7\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}4&\color{orangered}{ 3 }&-7&-16&7\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 3 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&3&-7&-16&7\\& & \color{blue}{12} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 12 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrr}4&3&\color{orangered}{ -7 }&-16&7\\& & \color{orangered}{12} & & \\ \hline &3&\color{orangered}{5}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&3&-7&-16&7\\& & 12& \color{blue}{20} & \\ \hline &3&\color{blue}{5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 20 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrr}4&3&-7&\color{orangered}{ -16 }&7\\& & 12& \color{orangered}{20} & \\ \hline &3&5&\color{orangered}{4}& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&3&-7&-16&7\\& & 12& 20& \color{blue}{16} \\ \hline &3&5&\color{blue}{4}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 16 } = \color{orangered}{ 23 } $
$$ \begin{array}{c|rrrr}4&3&-7&-16&\color{orangered}{ 7 }\\& & 12& 20& \color{orangered}{16} \\ \hline &\color{blue}{3}&\color{blue}{5}&\color{blue}{4}&\color{orangered}{23} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 23 }\right) $.