Check if $ x-3 $ is a factor of $ x^{3}+4x^{2}+3x-8 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&1&4&3&-8\\& & 3& 21& \color{black}{72} \\ \hline &\color{blue}{1}&\color{blue}{7}&\color{blue}{24}&\color{orangered}{64} \end{array} $$Because the remainder $ \left( \color{red}{ 64 } \right) $ is not zero, we conclude that the $ x-3 $ is not a factor of $ x^{3}+4x^{2}+3x-8$.
Explanation
First we need to create a synthetic division table.
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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&4&3&-8\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 1 }&4&3&-8\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&4&3&-8\\& & \color{blue}{3} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 3 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrr}3&1&\color{orangered}{ 4 }&3&-8\\& & \color{orangered}{3} & & \\ \hline &1&\color{orangered}{7}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&4&3&-8\\& & 3& \color{blue}{21} & \\ \hline &1&\color{blue}{7}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 21 } = \color{orangered}{ 24 } $
$$ \begin{array}{c|rrrr}3&1&4&\color{orangered}{ 3 }&-8\\& & 3& \color{orangered}{21} & \\ \hline &1&7&\color{orangered}{24}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 24 } = \color{blue}{ 72 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&4&3&-8\\& & 3& 21& \color{blue}{72} \\ \hline &1&7&\color{blue}{24}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 72 } = \color{orangered}{ 64 } $
$$ \begin{array}{c|rrrr}3&1&4&3&\color{orangered}{ -8 }\\& & 3& 21& \color{orangered}{72} \\ \hline &\color{blue}{1}&\color{blue}{7}&\color{blue}{24}&\color{orangered}{64} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 64 }\right)$.