Find remainder when $ 4x^{3}+10x^{2}-6x-20 $ is divided by $ x-1 $.
The synthetic division table is:
$$ \begin{array}{c|rrrr}1&4&10&-6&-20\\& & 4& 14& \color{black}{8} \\ \hline &\color{blue}{4}&\color{blue}{14}&\color{blue}{8}&\color{orangered}{-12} \end{array} $$The remainder when $ 4x^{3}+10x^{2}-6x-20 $ is divided by $ x-1 $ is $ \, \color{red}{ -12 } $.
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 -1 = 0 $ ( $ x = \color{blue}{ 1 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{1}&4&10&-6&-20\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}1&\color{orangered}{ 4 }&10&-6&-20\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 4 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&4&10&-6&-20\\& & \color{blue}{4} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ 4 } = \color{orangered}{ 14 } $
$$ \begin{array}{c|rrrr}1&4&\color{orangered}{ 10 }&-6&-20\\& & \color{orangered}{4} & & \\ \hline &4&\color{orangered}{14}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 14 } = \color{blue}{ 14 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&4&10&-6&-20\\& & 4& \color{blue}{14} & \\ \hline &4&\color{blue}{14}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 14 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrr}1&4&10&\color{orangered}{ -6 }&-20\\& & 4& \color{orangered}{14} & \\ \hline &4&14&\color{orangered}{8}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 8 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&4&10&-6&-20\\& & 4& 14& \color{blue}{8} \\ \hline &4&14&\color{blue}{8}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -20 } + \color{orangered}{ 8 } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrr}1&4&10&-6&\color{orangered}{ -20 }\\& & 4& 14& \color{orangered}{8} \\ \hline &\color{blue}{4}&\color{blue}{14}&\color{blue}{8}&\color{orangered}{-12} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -12 }\right) $.