Find remainder when $ 4x^{4}-3x^{3}-2x^{2}-8x+10 $ is divided by $ x-3 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&4&-3&-2&-8&10\\& & 12& 27& 75& \color{black}{201} \\ \hline &\color{blue}{4}&\color{blue}{9}&\color{blue}{25}&\color{blue}{67}&\color{orangered}{211} \end{array} $$The remainder when $ 4x^{4}-3x^{3}-2x^{2}-8x+10 $ is divided by $ x-3 $ is $ \, \color{red}{ 211 } $.
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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&4&-3&-2&-8&10\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 4 }&-3&-2&-8&10\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 4 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&4&-3&-2&-8&10\\& & \color{blue}{12} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 12 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrr}3&4&\color{orangered}{ -3 }&-2&-8&10\\& & \color{orangered}{12} & & & \\ \hline &4&\color{orangered}{9}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 9 } = \color{blue}{ 27 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&4&-3&-2&-8&10\\& & 12& \color{blue}{27} & & \\ \hline &4&\color{blue}{9}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 27 } = \color{orangered}{ 25 } $
$$ \begin{array}{c|rrrrr}3&4&-3&\color{orangered}{ -2 }&-8&10\\& & 12& \color{orangered}{27} & & \\ \hline &4&9&\color{orangered}{25}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 25 } = \color{blue}{ 75 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&4&-3&-2&-8&10\\& & 12& 27& \color{blue}{75} & \\ \hline &4&9&\color{blue}{25}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 75 } = \color{orangered}{ 67 } $
$$ \begin{array}{c|rrrrr}3&4&-3&-2&\color{orangered}{ -8 }&10\\& & 12& 27& \color{orangered}{75} & \\ \hline &4&9&25&\color{orangered}{67}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 67 } = \color{blue}{ 201 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&4&-3&-2&-8&10\\& & 12& 27& 75& \color{blue}{201} \\ \hline &4&9&25&\color{blue}{67}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ 201 } = \color{orangered}{ 211 } $
$$ \begin{array}{c|rrrrr}3&4&-3&-2&-8&\color{orangered}{ 10 }\\& & 12& 27& 75& \color{orangered}{201} \\ \hline &\color{blue}{4}&\color{blue}{9}&\color{blue}{25}&\color{blue}{67}&\color{orangered}{211} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 211 }\right) $.