Find remainder when $ x^{5}+2x^{4}-23x^{3}+34x^{2}-32x+6 $ is divided by $ x-3 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}3&1&2&-23&34&-32&6\\& & 3& 15& -24& 30& \color{black}{-6} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{-8}&\color{blue}{10}&\color{blue}{-2}&\color{orangered}{0} \end{array} $$The remainder when $ x^{5}+2x^{4}-23x^{3}+34x^{2}-32x+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.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&2&-23&34&-32&6\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}3&\color{orangered}{ 1 }&2&-23&34&-32&6\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&2&-23&34&-32&6\\& & \color{blue}{3} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 3 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrrr}3&1&\color{orangered}{ 2 }&-23&34&-32&6\\& & \color{orangered}{3} & & & & \\ \hline &1&\color{orangered}{5}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 5 } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&2&-23&34&-32&6\\& & 3& \color{blue}{15} & & & \\ \hline &1&\color{blue}{5}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -23 } + \color{orangered}{ 15 } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrrr}3&1&2&\color{orangered}{ -23 }&34&-32&6\\& & 3& \color{orangered}{15} & & & \\ \hline &1&5&\color{orangered}{-8}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ -24 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&2&-23&34&-32&6\\& & 3& 15& \color{blue}{-24} & & \\ \hline &1&5&\color{blue}{-8}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 34 } + \color{orangered}{ \left( -24 \right) } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrrr}3&1&2&-23&\color{orangered}{ 34 }&-32&6\\& & 3& 15& \color{orangered}{-24} & & \\ \hline &1&5&-8&\color{orangered}{10}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 10 } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&2&-23&34&-32&6\\& & 3& 15& -24& \color{blue}{30} & \\ \hline &1&5&-8&\color{blue}{10}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -32 } + \color{orangered}{ 30 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrrr}3&1&2&-23&34&\color{orangered}{ -32 }&6\\& & 3& 15& -24& \color{orangered}{30} & \\ \hline &1&5&-8&10&\color{orangered}{-2}& \end{array} $$Step 10 : 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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&2&-23&34&-32&6\\& & 3& 15& -24& 30& \color{blue}{-6} \\ \hline &1&5&-8&10&\color{blue}{-2}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrr}3&1&2&-23&34&-32&\color{orangered}{ 6 }\\& & 3& 15& -24& 30& \color{orangered}{-6} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{-8}&\color{blue}{10}&\color{blue}{-2}&\color{orangered}{0} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right) $.