Find remainder when $ 4x^{3}-16x^{2}+x-4 $ is divided by $ x-5 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}5&4&-16&1&-4\\& & 20& 20& \color{black}{105} \\ \hline &\color{blue}{4}&\color{blue}{4}&\color{blue}{21}&\color{orangered}{101} \end{array} $$The remainder when $ 4x^{3}-16x^{2}+x-4 $ is divided by $ x-5 $ is $ \, \color{red}{ 101 } $.
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.
$$ \begin{array}{c|rrrr}\color{blue}{5}&4&-16&1&-4\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}5&\color{orangered}{ 4 }&-16&1&-4\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 4 } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&4&-16&1&-4\\& & \color{blue}{20} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 20 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrr}5&4&\color{orangered}{ -16 }&1&-4\\& & \color{orangered}{20} & & \\ \hline &4&\color{orangered}{4}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 4 } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&4&-16&1&-4\\& & 20& \color{blue}{20} & \\ \hline &4&\color{blue}{4}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 20 } = \color{orangered}{ 21 } $
$$ \begin{array}{c|rrrr}5&4&-16&\color{orangered}{ 1 }&-4\\& & 20& \color{orangered}{20} & \\ \hline &4&4&\color{orangered}{21}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 21 } = \color{blue}{ 105 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&4&-16&1&-4\\& & 20& 20& \color{blue}{105} \\ \hline &4&4&\color{blue}{21}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 105 } = \color{orangered}{ 101 } $
$$ \begin{array}{c|rrrr}5&4&-16&1&\color{orangered}{ -4 }\\& & 20& 20& \color{orangered}{105} \\ \hline &\color{blue}{4}&\color{blue}{4}&\color{blue}{21}&\color{orangered}{101} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 101 }\right) $.