Divide using synthetic division $$ ( 2x^{5}-x^{3} ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}3&2&0&-1&0&0&0\\& & 6& 18& 51& 153& \color{black}{459} \\ \hline &\color{blue}{2}&\color{blue}{6}&\color{blue}{17}&\color{blue}{51}&\color{blue}{153}&\color{orangered}{459} \end{array} $$The solution is:
$$ \frac{ 2x^{5}-x^{3} }{ x-3 } = \color{blue}{2x^{4}+6x^{3}+17x^{2}+51x+153} ~+~ \frac{ \color{red}{ 459 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&2&0&-1&0&0&0\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}3&\color{orangered}{ 2 }&0&-1&0&0&0\\& & & & & & \\ \hline &\color{orangered}{2}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&2&0&-1&0&0&0\\& & \color{blue}{6} & & & & \\ \hline &\color{blue}{2}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 6 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrrr}3&2&\color{orangered}{ 0 }&-1&0&0&0\\& & \color{orangered}{6} & & & & \\ \hline &2&\color{orangered}{6}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 6 } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&2&0&-1&0&0&0\\& & 6& \color{blue}{18} & & & \\ \hline &2&\color{blue}{6}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 18 } = \color{orangered}{ 17 } $
$$ \begin{array}{c|rrrrrr}3&2&0&\color{orangered}{ -1 }&0&0&0\\& & 6& \color{orangered}{18} & & & \\ \hline &2&6&\color{orangered}{17}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 17 } = \color{blue}{ 51 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&2&0&-1&0&0&0\\& & 6& 18& \color{blue}{51} & & \\ \hline &2&6&\color{blue}{17}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 51 } = \color{orangered}{ 51 } $
$$ \begin{array}{c|rrrrrr}3&2&0&-1&\color{orangered}{ 0 }&0&0\\& & 6& 18& \color{orangered}{51} & & \\ \hline &2&6&17&\color{orangered}{51}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 51 } = \color{blue}{ 153 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&2&0&-1&0&0&0\\& & 6& 18& 51& \color{blue}{153} & \\ \hline &2&6&17&\color{blue}{51}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 153 } = \color{orangered}{ 153 } $
$$ \begin{array}{c|rrrrrr}3&2&0&-1&0&\color{orangered}{ 0 }&0\\& & 6& 18& 51& \color{orangered}{153} & \\ \hline &2&6&17&51&\color{orangered}{153}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 153 } = \color{blue}{ 459 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&2&0&-1&0&0&0\\& & 6& 18& 51& 153& \color{blue}{459} \\ \hline &2&6&17&51&\color{blue}{153}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 459 } = \color{orangered}{ 459 } $
$$ \begin{array}{c|rrrrrr}3&2&0&-1&0&0&\color{orangered}{ 0 }\\& & 6& 18& 51& 153& \color{orangered}{459} \\ \hline &\color{blue}{2}&\color{blue}{6}&\color{blue}{17}&\color{blue}{51}&\color{blue}{153}&\color{orangered}{459} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{4}+6x^{3}+17x^{2}+51x+153 } $ with a remainder of $ \color{red}{ 459 } $.