Find remainder when $ 8x^{4}-36x^{3}-24x^{2}+20x $ is divided by $ x-5 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}5&8&-36&-24&20&0\\& & 40& 20& -20& \color{black}{0} \\ \hline &\color{blue}{8}&\color{blue}{4}&\color{blue}{-4}&\color{blue}{0}&\color{orangered}{0} \end{array} $$The remainder when $ 8x^{4}-36x^{3}-24x^{2}+20x $ is divided by $ x-5 $ 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 -5 = 0 $ ( $ x = \color{blue}{ 5 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&8&-36&-24&20&0\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}5&\color{orangered}{ 8 }&-36&-24&20&0\\& & & & & \\ \hline &\color{orangered}{8}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 8 } = \color{blue}{ 40 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&8&-36&-24&20&0\\& & \color{blue}{40} & & & \\ \hline &\color{blue}{8}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -36 } + \color{orangered}{ 40 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}5&8&\color{orangered}{ -36 }&-24&20&0\\& & \color{orangered}{40} & & & \\ \hline &8&\color{orangered}{4}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 4 } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&8&-36&-24&20&0\\& & 40& \color{blue}{20} & & \\ \hline &8&\color{blue}{4}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -24 } + \color{orangered}{ 20 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrr}5&8&-36&\color{orangered}{ -24 }&20&0\\& & 40& \color{orangered}{20} & & \\ \hline &8&4&\color{orangered}{-4}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&8&-36&-24&20&0\\& & 40& 20& \color{blue}{-20} & \\ \hline &8&4&\color{blue}{-4}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 20 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}5&8&-36&-24&\color{orangered}{ 20 }&0\\& & 40& 20& \color{orangered}{-20} & \\ \hline &8&4&-4&\color{orangered}{0}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&8&-36&-24&20&0\\& & 40& 20& -20& \color{blue}{0} \\ \hline &8&4&-4&\color{blue}{0}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}5&8&-36&-24&20&\color{orangered}{ 0 }\\& & 40& 20& -20& \color{orangered}{0} \\ \hline &\color{blue}{8}&\color{blue}{4}&\color{blue}{-4}&\color{blue}{0}&\color{orangered}{0} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right) $.