Divide using synthetic division $$ ( 2x^{4}-5x^{3}+7x^{2}-3x+1 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&2&-5&7&-3&1\\& & 6& 3& 30& \color{black}{81} \\ \hline &\color{blue}{2}&\color{blue}{1}&\color{blue}{10}&\color{blue}{27}&\color{orangered}{82} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-5x^{3}+7x^{2}-3x+1 }{ x-3 } = \color{blue}{2x^{3}+x^{2}+10x+27} ~+~ \frac{ \color{red}{ 82 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-5&7&-3&1\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 2 }&-5&7&-3&1\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-5&7&-3&1\\& & \color{blue}{6} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 6 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}3&2&\color{orangered}{ -5 }&7&-3&1\\& & \color{orangered}{6} & & & \\ \hline &2&\color{orangered}{1}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 1 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-5&7&-3&1\\& & 6& \color{blue}{3} & & \\ \hline &2&\color{blue}{1}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 3 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrr}3&2&-5&\color{orangered}{ 7 }&-3&1\\& & 6& \color{orangered}{3} & & \\ \hline &2&1&\color{orangered}{10}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 10 } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-5&7&-3&1\\& & 6& 3& \color{blue}{30} & \\ \hline &2&1&\color{blue}{10}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 30 } = \color{orangered}{ 27 } $
$$ \begin{array}{c|rrrrr}3&2&-5&7&\color{orangered}{ -3 }&1\\& & 6& 3& \color{orangered}{30} & \\ \hline &2&1&10&\color{orangered}{27}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 27 } = \color{blue}{ 81 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-5&7&-3&1\\& & 6& 3& 30& \color{blue}{81} \\ \hline &2&1&10&\color{blue}{27}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 81 } = \color{orangered}{ 82 } $
$$ \begin{array}{c|rrrrr}3&2&-5&7&-3&\color{orangered}{ 1 }\\& & 6& 3& 30& \color{orangered}{81} \\ \hline &\color{blue}{2}&\color{blue}{1}&\color{blue}{10}&\color{blue}{27}&\color{orangered}{82} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}+x^{2}+10x+27 } $ with a remainder of $ \color{red}{ 82 } $.