Divide using synthetic division $$ ( 5x^{5}-10x^{4}-15x^{3}+9x^{2}-19x+17 ) \div ( x-3 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}3&5&-10&-15&9&-19&17\\& & 15& 15& 0& 27& \color{black}{24} \\ \hline &\color{blue}{5}&\color{blue}{5}&\color{blue}{0}&\color{blue}{9}&\color{blue}{8}&\color{orangered}{41} \end{array} $$The solution is:
$$ \dfrac{ 5x^{5}-10x^{4}-15x^{3}+9x^{2}-19x+17 }{ x-3 } = \color{blue}{5x^{4}+5x^{3}+9x+8} ~+~ \dfrac{ \color{red}{ 41 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&5&-10&-15&9&-19&17\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}3&\color{orangered}{ 5 }&-10&-15&9&-19&17\\& & & & & & \\ \hline &\color{orangered}{5}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&5&-10&-15&9&-19&17\\& & \color{blue}{15} & & & & \\ \hline &\color{blue}{5}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 15 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrrr}3&5&\color{orangered}{ -10 }&-15&9&-19&17\\& & \color{orangered}{15} & & & & \\ \hline &5&\color{orangered}{5}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 5 } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&5&-10&-15&9&-19&17\\& & 15& \color{blue}{15} & & & \\ \hline &5&\color{blue}{5}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ 15 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrr}3&5&-10&\color{orangered}{ -15 }&9&-19&17\\& & 15& \color{orangered}{15} & & & \\ \hline &5&5&\color{orangered}{0}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&5&-10&-15&9&-19&17\\& & 15& 15& \color{blue}{0} & & \\ \hline &5&5&\color{blue}{0}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ 0 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrrr}3&5&-10&-15&\color{orangered}{ 9 }&-19&17\\& & 15& 15& \color{orangered}{0} & & \\ \hline &5&5&0&\color{orangered}{9}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 9 } = \color{blue}{ 27 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&5&-10&-15&9&-19&17\\& & 15& 15& 0& \color{blue}{27} & \\ \hline &5&5&0&\color{blue}{9}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -19 } + \color{orangered}{ 27 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrrr}3&5&-10&-15&9&\color{orangered}{ -19 }&17\\& & 15& 15& 0& \color{orangered}{27} & \\ \hline &5&5&0&9&\color{orangered}{8}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 8 } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&5&-10&-15&9&-19&17\\& & 15& 15& 0& 27& \color{blue}{24} \\ \hline &5&5&0&9&\color{blue}{8}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 17 } + \color{orangered}{ 24 } = \color{orangered}{ 41 } $
$$ \begin{array}{c|rrrrrr}3&5&-10&-15&9&-19&\color{orangered}{ 17 }\\& & 15& 15& 0& 27& \color{orangered}{24} \\ \hline &\color{blue}{5}&\color{blue}{5}&\color{blue}{0}&\color{blue}{9}&\color{blue}{8}&\color{orangered}{41} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{4}+5x^{3}+9x+8 } $ with a remainder of $ \color{red}{ 41 } $.