Divide using synthetic division $$ ( 3x^{5}-29x^{3}+20x+1 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}3&3&0&-29&0&20&1\\& & 9& 27& -6& -18& \color{black}{6} \\ \hline &\color{blue}{3}&\color{blue}{9}&\color{blue}{-2}&\color{blue}{-6}&\color{blue}{2}&\color{orangered}{7} \end{array} $$The solution is:
$$ \frac{ 3x^{5}-29x^{3}+20x+1 }{ x-3 } = \color{blue}{3x^{4}+9x^{3}-2x^{2}-6x+2} ~+~ \frac{ \color{red}{ 7 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&3&0&-29&0&20&1\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}3&\color{orangered}{ 3 }&0&-29&0&20&1\\& & & & & & \\ \hline &\color{orangered}{3}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 3 } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&3&0&-29&0&20&1\\& & \color{blue}{9} & & & & \\ \hline &\color{blue}{3}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 9 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrrr}3&3&\color{orangered}{ 0 }&-29&0&20&1\\& & \color{orangered}{9} & & & & \\ \hline &3&\color{orangered}{9}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 9 } = \color{blue}{ 27 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&3&0&-29&0&20&1\\& & 9& \color{blue}{27} & & & \\ \hline &3&\color{blue}{9}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -29 } + \color{orangered}{ 27 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrrr}3&3&0&\color{orangered}{ -29 }&0&20&1\\& & 9& \color{orangered}{27} & & & \\ \hline &3&9&\color{orangered}{-2}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&3&0&-29&0&20&1\\& & 9& 27& \color{blue}{-6} & & \\ \hline &3&9&\color{blue}{-2}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrrr}3&3&0&-29&\color{orangered}{ 0 }&20&1\\& & 9& 27& \color{orangered}{-6} & & \\ \hline &3&9&-2&\color{orangered}{-6}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ -18 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&3&0&-29&0&20&1\\& & 9& 27& -6& \color{blue}{-18} & \\ \hline &3&9&-2&\color{blue}{-6}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 20 } + \color{orangered}{ \left( -18 \right) } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrrr}3&3&0&-29&0&\color{orangered}{ 20 }&1\\& & 9& 27& -6& \color{orangered}{-18} & \\ \hline &3&9&-2&-6&\color{orangered}{2}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 2 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&3&0&-29&0&20&1\\& & 9& 27& -6& -18& \color{blue}{6} \\ \hline &3&9&-2&-6&\color{blue}{2}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 6 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrrrr}3&3&0&-29&0&20&\color{orangered}{ 1 }\\& & 9& 27& -6& -18& \color{orangered}{6} \\ \hline &\color{blue}{3}&\color{blue}{9}&\color{blue}{-2}&\color{blue}{-6}&\color{blue}{2}&\color{orangered}{7} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{4}+9x^{3}-2x^{2}-6x+2 } $ with a remainder of $ \color{red}{ 7 } $.