Divide using synthetic division $$ ( 2x^{4}-15x^{3}+19x^{2}-9x+18 ) \div ( x-6 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}6&2&-15&19&-9&18\\& & 12& -18& 6& \color{black}{-18} \\ \hline &\color{blue}{2}&\color{blue}{-3}&\color{blue}{1}&\color{blue}{-3}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-15x^{3}+19x^{2}-9x+18 }{ x-6 } = \color{blue}{2x^{3}-3x^{2}+x-3} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -6 = 0 $ ( $ x = \color{blue}{ 6 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{6}&2&-15&19&-9&18\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}6&\color{orangered}{ 2 }&-15&19&-9&18\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 2 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{6}&2&-15&19&-9&18\\& & \color{blue}{12} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ 12 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}6&2&\color{orangered}{ -15 }&19&-9&18\\& & \color{orangered}{12} & & & \\ \hline &2&\color{orangered}{-3}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ -18 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{6}&2&-15&19&-9&18\\& & 12& \color{blue}{-18} & & \\ \hline &2&\color{blue}{-3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 19 } + \color{orangered}{ \left( -18 \right) } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}6&2&-15&\color{orangered}{ 19 }&-9&18\\& & 12& \color{orangered}{-18} & & \\ \hline &2&-3&\color{orangered}{1}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 1 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{6}&2&-15&19&-9&18\\& & 12& -18& \color{blue}{6} & \\ \hline &2&-3&\color{blue}{1}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -9 } + \color{orangered}{ 6 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}6&2&-15&19&\color{orangered}{ -9 }&18\\& & 12& -18& \color{orangered}{6} & \\ \hline &2&-3&1&\color{orangered}{-3}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ -18 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{6}&2&-15&19&-9&18\\& & 12& -18& 6& \color{blue}{-18} \\ \hline &2&-3&1&\color{blue}{-3}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 18 } + \color{orangered}{ \left( -18 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}6&2&-15&19&-9&\color{orangered}{ 18 }\\& & 12& -18& 6& \color{orangered}{-18} \\ \hline &\color{blue}{2}&\color{blue}{-3}&\color{blue}{1}&\color{blue}{-3}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}-3x^{2}+x-3 } $ with a remainder of $ \color{red}{ 0 } $.