Find remainder when $ 7x^{2}-23x+6 $ is divided by $ x-3 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrr}3&7&-23&6\\& & 21& \color{black}{-6} \\ \hline &\color{blue}{7}&\color{blue}{-2}&\color{orangered}{0} \end{array} $$The remainder when $ 7x^{2}-23x+6 $ is divided by $ x-3 $ is $ \, \color{red}{ 0 } $.
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.
$$ \begin{array}{c|rrr}\color{blue}{3}&7&-23&6\\& & & \\ \hline &&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrr}3&\color{orangered}{ 7 }&-23&6\\& & & \\ \hline &\color{orangered}{7}&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 7 } = \color{blue}{ 21 } $.
$$ \begin{array}{c|rrr}\color{blue}{3}&7&-23&6\\& & \color{blue}{21} & \\ \hline &\color{blue}{7}&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -23 } + \color{orangered}{ 21 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrr}3&7&\color{orangered}{ -23 }&6\\& & \color{orangered}{21} & \\ \hline &7&\color{orangered}{-2}& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrr}\color{blue}{3}&7&-23&6\\& & 21& \color{blue}{-6} \\ \hline &7&\color{blue}{-2}& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrr}3&7&-23&\color{orangered}{ 6 }\\& & 21& \color{orangered}{-6} \\ \hline &\color{blue}{7}&\color{blue}{-2}&\color{orangered}{0} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right) $.