Find remainder when $ 7x^{5}-20x^{4}-16x^{2}+12x+19 $ is divided by $ x-3 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}3&7&-20&0&-16&12&19\\& & 21& 3& 9& -21& \color{black}{-27} \\ \hline &\color{blue}{7}&\color{blue}{1}&\color{blue}{3}&\color{blue}{-7}&\color{blue}{-9}&\color{orangered}{-8} \end{array} $$The remainder when $ 7x^{5}-20x^{4}-16x^{2}+12x+19 $ is divided by $ x-3 $ is $ \, \color{red}{ -8 } $.
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.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&7&-20&0&-16&12&19\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}3&\color{orangered}{ 7 }&-20&0&-16&12&19\\& & & & & & \\ \hline &\color{orangered}{7}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&7&-20&0&-16&12&19\\& & \color{blue}{21} & & & & \\ \hline &\color{blue}{7}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -20 } + \color{orangered}{ 21 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrrr}3&7&\color{orangered}{ -20 }&0&-16&12&19\\& & \color{orangered}{21} & & & & \\ \hline &7&\color{orangered}{1}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 1 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&7&-20&0&-16&12&19\\& & 21& \color{blue}{3} & & & \\ \hline &7&\color{blue}{1}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 3 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrrr}3&7&-20&\color{orangered}{ 0 }&-16&12&19\\& & 21& \color{orangered}{3} & & & \\ \hline &7&1&\color{orangered}{3}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 3 } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&7&-20&0&-16&12&19\\& & 21& 3& \color{blue}{9} & & \\ \hline &7&1&\color{blue}{3}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 9 } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrrrr}3&7&-20&0&\color{orangered}{ -16 }&12&19\\& & 21& 3& \color{orangered}{9} & & \\ \hline &7&1&3&\color{orangered}{-7}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ -21 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&7&-20&0&-16&12&19\\& & 21& 3& 9& \color{blue}{-21} & \\ \hline &7&1&3&\color{blue}{-7}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -21 \right) } = \color{orangered}{ -9 } $
$$ \begin{array}{c|rrrrrr}3&7&-20&0&-16&\color{orangered}{ 12 }&19\\& & 21& 3& 9& \color{orangered}{-21} & \\ \hline &7&1&3&-7&\color{orangered}{-9}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -9 \right) } = \color{blue}{ -27 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&7&-20&0&-16&12&19\\& & 21& 3& 9& -21& \color{blue}{-27} \\ \hline &7&1&3&-7&\color{blue}{-9}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 19 } + \color{orangered}{ \left( -27 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrrr}3&7&-20&0&-16&12&\color{orangered}{ 19 }\\& & 21& 3& 9& -21& \color{orangered}{-27} \\ \hline &\color{blue}{7}&\color{blue}{1}&\color{blue}{3}&\color{blue}{-7}&\color{blue}{-9}&\color{orangered}{-8} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -8 }\right) $.