Divide using synthetic division $$ ( 2x^{5}-6x^{4}+2x^{3}+4x^{2}-5x+3 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}3&2&-6&2&4&-5&3\\& & 6& 0& 6& 30& \color{black}{75} \\ \hline &\color{blue}{2}&\color{blue}{0}&\color{blue}{2}&\color{blue}{10}&\color{blue}{25}&\color{orangered}{78} \end{array} $$The solution is:
$$ \frac{ 2x^{5}-6x^{4}+2x^{3}+4x^{2}-5x+3 }{ x-3 } = \color{blue}{2x^{4}+2x^{2}+10x+25} ~+~ \frac{ \color{red}{ 78 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&2&-6&2&4&-5&3\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}3&\color{orangered}{ 2 }&-6&2&4&-5&3\\& & & & & & \\ \hline &\color{orangered}{2}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&2&-6&2&4&-5&3\\& & \color{blue}{6} & & & & \\ \hline &\color{blue}{2}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 6 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrr}3&2&\color{orangered}{ -6 }&2&4&-5&3\\& & \color{orangered}{6} & & & & \\ \hline &2&\color{orangered}{0}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&2&-6&2&4&-5&3\\& & 6& \color{blue}{0} & & & \\ \hline &2&\color{blue}{0}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 0 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrrr}3&2&-6&\color{orangered}{ 2 }&4&-5&3\\& & 6& \color{orangered}{0} & & & \\ \hline &2&0&\color{orangered}{2}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 2 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&2&-6&2&4&-5&3\\& & 6& 0& \color{blue}{6} & & \\ \hline &2&0&\color{blue}{2}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 6 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrrr}3&2&-6&2&\color{orangered}{ 4 }&-5&3\\& & 6& 0& \color{orangered}{6} & & \\ \hline &2&0&2&\color{orangered}{10}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 10 } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&2&-6&2&4&-5&3\\& & 6& 0& 6& \color{blue}{30} & \\ \hline &2&0&2&\color{blue}{10}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 30 } = \color{orangered}{ 25 } $
$$ \begin{array}{c|rrrrrr}3&2&-6&2&4&\color{orangered}{ -5 }&3\\& & 6& 0& 6& \color{orangered}{30} & \\ \hline &2&0&2&10&\color{orangered}{25}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 25 } = \color{blue}{ 75 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&2&-6&2&4&-5&3\\& & 6& 0& 6& 30& \color{blue}{75} \\ \hline &2&0&2&10&\color{blue}{25}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 75 } = \color{orangered}{ 78 } $
$$ \begin{array}{c|rrrrrr}3&2&-6&2&4&-5&\color{orangered}{ 3 }\\& & 6& 0& 6& 30& \color{orangered}{75} \\ \hline &\color{blue}{2}&\color{blue}{0}&\color{blue}{2}&\color{blue}{10}&\color{blue}{25}&\color{orangered}{78} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{4}+2x^{2}+10x+25 } $ with a remainder of $ \color{red}{ 78 } $.