Divide using synthetic division $$ ( 5x^{5}-15x^{3}+9x^{2}-19x+7 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}3&5&0&-15&9&-19&7\\& & 15& 45& 90& 297& \color{black}{834} \\ \hline &\color{blue}{5}&\color{blue}{15}&\color{blue}{30}&\color{blue}{99}&\color{blue}{278}&\color{orangered}{841} \end{array} $$The solution is:
$$ \frac{ 5x^{5}-15x^{3}+9x^{2}-19x+7 }{ x-3 } = \color{blue}{5x^{4}+15x^{3}+30x^{2}+99x+278} ~+~ \frac{ \color{red}{ 841 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&5&0&-15&9&-19&7\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}3&\color{orangered}{ 5 }&0&-15&9&-19&7\\& & & & & & \\ \hline &\color{orangered}{5}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 5 } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&5&0&-15&9&-19&7\\& & \color{blue}{15} & & & & \\ \hline &\color{blue}{5}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 15 } = \color{orangered}{ 15 } $
$$ \begin{array}{c|rrrrrr}3&5&\color{orangered}{ 0 }&-15&9&-19&7\\& & \color{orangered}{15} & & & & \\ \hline &5&\color{orangered}{15}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 15 } = \color{blue}{ 45 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&5&0&-15&9&-19&7\\& & 15& \color{blue}{45} & & & \\ \hline &5&\color{blue}{15}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ 45 } = \color{orangered}{ 30 } $
$$ \begin{array}{c|rrrrrr}3&5&0&\color{orangered}{ -15 }&9&-19&7\\& & 15& \color{orangered}{45} & & & \\ \hline &5&15&\color{orangered}{30}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 30 } = \color{blue}{ 90 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&5&0&-15&9&-19&7\\& & 15& 45& \color{blue}{90} & & \\ \hline &5&15&\color{blue}{30}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ 90 } = \color{orangered}{ 99 } $
$$ \begin{array}{c|rrrrrr}3&5&0&-15&\color{orangered}{ 9 }&-19&7\\& & 15& 45& \color{orangered}{90} & & \\ \hline &5&15&30&\color{orangered}{99}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 99 } = \color{blue}{ 297 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&5&0&-15&9&-19&7\\& & 15& 45& 90& \color{blue}{297} & \\ \hline &5&15&30&\color{blue}{99}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -19 } + \color{orangered}{ 297 } = \color{orangered}{ 278 } $
$$ \begin{array}{c|rrrrrr}3&5&0&-15&9&\color{orangered}{ -19 }&7\\& & 15& 45& 90& \color{orangered}{297} & \\ \hline &5&15&30&99&\color{orangered}{278}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 278 } = \color{blue}{ 834 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&5&0&-15&9&-19&7\\& & 15& 45& 90& 297& \color{blue}{834} \\ \hline &5&15&30&99&\color{blue}{278}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 834 } = \color{orangered}{ 841 } $
$$ \begin{array}{c|rrrrrr}3&5&0&-15&9&-19&\color{orangered}{ 7 }\\& & 15& 45& 90& 297& \color{orangered}{834} \\ \hline &\color{blue}{5}&\color{blue}{15}&\color{blue}{30}&\color{blue}{99}&\color{blue}{278}&\color{orangered}{841} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{4}+15x^{3}+30x^{2}+99x+278 } $ with a remainder of $ \color{red}{ 841 } $.