Find remainder when $ x^{3}+4x^{2}+x-4 $ is divided by $ x-3 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&1&4&1&-4\\& & 3& 21& \color{black}{66} \\ \hline &\color{blue}{1}&\color{blue}{7}&\color{blue}{22}&\color{orangered}{62} \end{array} $$The remainder when $ x^{3}+4x^{2}+x-4 $ is divided by $ x-3 $ is $ \, \color{red}{ 62 } $.
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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&4&1&-4\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 1 }&4&1&-4\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&4&1&-4\\& & \color{blue}{3} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 3 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrr}3&1&\color{orangered}{ 4 }&1&-4\\& & \color{orangered}{3} & & \\ \hline &1&\color{orangered}{7}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 7 } = \color{blue}{ 21 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&4&1&-4\\& & 3& \color{blue}{21} & \\ \hline &1&\color{blue}{7}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 21 } = \color{orangered}{ 22 } $
$$ \begin{array}{c|rrrr}3&1&4&\color{orangered}{ 1 }&-4\\& & 3& \color{orangered}{21} & \\ \hline &1&7&\color{orangered}{22}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 22 } = \color{blue}{ 66 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&4&1&-4\\& & 3& 21& \color{blue}{66} \\ \hline &1&7&\color{blue}{22}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 66 } = \color{orangered}{ 62 } $
$$ \begin{array}{c|rrrr}3&1&4&1&\color{orangered}{ -4 }\\& & 3& 21& \color{orangered}{66} \\ \hline &\color{blue}{1}&\color{blue}{7}&\color{blue}{22}&\color{orangered}{62} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 62 }\right) $.