Divide using synthetic division $$ ( 5x^{4}+x^{2}-8x+2 ) \div ( x-4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}4&5&0&1&-8&2\\& & 20& 80& 324& \color{black}{1264} \\ \hline &\color{blue}{5}&\color{blue}{20}&\color{blue}{81}&\color{blue}{316}&\color{orangered}{1266} \end{array} $$The solution is:
$$ \frac{ 5x^{4}+x^{2}-8x+2 }{ x-4 } = \color{blue}{5x^{3}+20x^{2}+81x+316} ~+~ \frac{ \color{red}{ 1266 } }{ x-4 } $$Explanation
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.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&5&0&1&-8&2\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}4&\color{orangered}{ 5 }&0&1&-8&2\\& & & & & \\ \hline &\color{orangered}{5}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&5&0&1&-8&2\\& & \color{blue}{20} & & & \\ \hline &\color{blue}{5}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 20 } = \color{orangered}{ 20 } $
$$ \begin{array}{c|rrrrr}4&5&\color{orangered}{ 0 }&1&-8&2\\& & \color{orangered}{20} & & & \\ \hline &5&\color{orangered}{20}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 20 } = \color{blue}{ 80 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&5&0&1&-8&2\\& & 20& \color{blue}{80} & & \\ \hline &5&\color{blue}{20}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 80 } = \color{orangered}{ 81 } $
$$ \begin{array}{c|rrrrr}4&5&0&\color{orangered}{ 1 }&-8&2\\& & 20& \color{orangered}{80} & & \\ \hline &5&20&\color{orangered}{81}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 81 } = \color{blue}{ 324 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&5&0&1&-8&2\\& & 20& 80& \color{blue}{324} & \\ \hline &5&20&\color{blue}{81}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 324 } = \color{orangered}{ 316 } $
$$ \begin{array}{c|rrrrr}4&5&0&1&\color{orangered}{ -8 }&2\\& & 20& 80& \color{orangered}{324} & \\ \hline &5&20&81&\color{orangered}{316}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 316 } = \color{blue}{ 1264 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&5&0&1&-8&2\\& & 20& 80& 324& \color{blue}{1264} \\ \hline &5&20&81&\color{blue}{316}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 1264 } = \color{orangered}{ 1266 } $
$$ \begin{array}{c|rrrrr}4&5&0&1&-8&\color{orangered}{ 2 }\\& & 20& 80& 324& \color{orangered}{1264} \\ \hline &\color{blue}{5}&\color{blue}{20}&\color{blue}{81}&\color{blue}{316}&\color{orangered}{1266} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{3}+20x^{2}+81x+316 } $ with a remainder of $ \color{red}{ 1266 } $.