Find remainder when $ 3x^{4}-13x^{3}+20x-16 $ is divided by $ x-4 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}4&3&-13&0&20&-16\\& & 12& -4& -16& \color{black}{16} \\ \hline &\color{blue}{3}&\color{blue}{-1}&\color{blue}{-4}&\color{blue}{4}&\color{orangered}{0} \end{array} $$The remainder when $ 3x^{4}-13x^{3}+20x-16 $ is divided by $ x-4 $ 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 -4 = 0 $ ( $ x = \color{blue}{ 4 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&3&-13&0&20&-16\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}4&\color{orangered}{ 3 }&-13&0&20&-16\\& & & & & \\ \hline &\color{orangered}{3}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 3 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&3&-13&0&20&-16\\& & \color{blue}{12} & & & \\ \hline &\color{blue}{3}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -13 } + \color{orangered}{ 12 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrr}4&3&\color{orangered}{ -13 }&0&20&-16\\& & \color{orangered}{12} & & & \\ \hline &3&\color{orangered}{-1}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&3&-13&0&20&-16\\& & 12& \color{blue}{-4} & & \\ \hline &3&\color{blue}{-1}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrr}4&3&-13&\color{orangered}{ 0 }&20&-16\\& & 12& \color{orangered}{-4} & & \\ \hline &3&-1&\color{orangered}{-4}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&3&-13&0&20&-16\\& & 12& -4& \color{blue}{-16} & \\ \hline &3&-1&\color{blue}{-4}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 20 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}4&3&-13&0&\color{orangered}{ 20 }&-16\\& & 12& -4& \color{orangered}{-16} & \\ \hline &3&-1&-4&\color{orangered}{4}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 4 } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&3&-13&0&20&-16\\& & 12& -4& -16& \color{blue}{16} \\ \hline &3&-1&-4&\color{blue}{4}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 16 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}4&3&-13&0&20&\color{orangered}{ -16 }\\& & 12& -4& -16& \color{orangered}{16} \\ \hline &\color{blue}{3}&\color{blue}{-1}&\color{blue}{-4}&\color{blue}{4}&\color{orangered}{0} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right) $.