Divide using synthetic division $$ ( 2x^{4}+8x^{3}+5x^{2}-5x+1 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&2&8&5&-5&1\\& & 6& 42& 141& \color{black}{408} \\ \hline &\color{blue}{2}&\color{blue}{14}&\color{blue}{47}&\color{blue}{136}&\color{orangered}{409} \end{array} $$The solution is:
$$ \frac{ 2x^{4}+8x^{3}+5x^{2}-5x+1 }{ x-3 } = \color{blue}{2x^{3}+14x^{2}+47x+136} ~+~ \frac{ \color{red}{ 409 } }{ x-3 } $$Explanation
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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&8&5&-5&1\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 2 }&8&5&-5&1\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 2 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&8&5&-5&1\\& & \color{blue}{6} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ 6 } = \color{orangered}{ 14 } $
$$ \begin{array}{c|rrrrr}3&2&\color{orangered}{ 8 }&5&-5&1\\& & \color{orangered}{6} & & & \\ \hline &2&\color{orangered}{14}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 14 } = \color{blue}{ 42 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&8&5&-5&1\\& & 6& \color{blue}{42} & & \\ \hline &2&\color{blue}{14}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 42 } = \color{orangered}{ 47 } $
$$ \begin{array}{c|rrrrr}3&2&8&\color{orangered}{ 5 }&-5&1\\& & 6& \color{orangered}{42} & & \\ \hline &2&14&\color{orangered}{47}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 47 } = \color{blue}{ 141 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&8&5&-5&1\\& & 6& 42& \color{blue}{141} & \\ \hline &2&14&\color{blue}{47}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 141 } = \color{orangered}{ 136 } $
$$ \begin{array}{c|rrrrr}3&2&8&5&\color{orangered}{ -5 }&1\\& & 6& 42& \color{orangered}{141} & \\ \hline &2&14&47&\color{orangered}{136}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 136 } = \color{blue}{ 408 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&8&5&-5&1\\& & 6& 42& 141& \color{blue}{408} \\ \hline &2&14&47&\color{blue}{136}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 408 } = \color{orangered}{ 409 } $
$$ \begin{array}{c|rrrrr}3&2&8&5&-5&\color{orangered}{ 1 }\\& & 6& 42& 141& \color{orangered}{408} \\ \hline &\color{blue}{2}&\color{blue}{14}&\color{blue}{47}&\color{blue}{136}&\color{orangered}{409} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}+14x^{2}+47x+136 } $ with a remainder of $ \color{red}{ 409 } $.