Find remainder when $ 5x^{3}+3x^{2}-8 $ is divided by $ x-4 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}4&5&3&0&-8\\& & 20& 92& \color{black}{368} \\ \hline &\color{blue}{5}&\color{blue}{23}&\color{blue}{92}&\color{orangered}{360} \end{array} $$The remainder when $ 5x^{3}+3x^{2}-8 $ is divided by $ x-4 $ is $ \, \color{red}{ 360 } $.
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.
$$ \begin{array}{c|rrrr}\color{blue}{4}&5&3&0&-8\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}4&\color{orangered}{ 5 }&3&0&-8\\& & & & \\ \hline &\color{orangered}{5}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 5 } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&5&3&0&-8\\& & \color{blue}{20} & & \\ \hline &\color{blue}{5}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 20 } = \color{orangered}{ 23 } $
$$ \begin{array}{c|rrrr}4&5&\color{orangered}{ 3 }&0&-8\\& & \color{orangered}{20} & & \\ \hline &5&\color{orangered}{23}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 23 } = \color{blue}{ 92 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&5&3&0&-8\\& & 20& \color{blue}{92} & \\ \hline &5&\color{blue}{23}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 92 } = \color{orangered}{ 92 } $
$$ \begin{array}{c|rrrr}4&5&3&\color{orangered}{ 0 }&-8\\& & 20& \color{orangered}{92} & \\ \hline &5&23&\color{orangered}{92}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 92 } = \color{blue}{ 368 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&5&3&0&-8\\& & 20& 92& \color{blue}{368} \\ \hline &5&23&\color{blue}{92}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 368 } = \color{orangered}{ 360 } $
$$ \begin{array}{c|rrrr}4&5&3&0&\color{orangered}{ -8 }\\& & 20& 92& \color{orangered}{368} \\ \hline &\color{blue}{5}&\color{blue}{23}&\color{blue}{92}&\color{orangered}{360} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 360 }\right) $.