Divide using synthetic division $$ ( 2x^{4}+6x^{3}-25x^{2}+19x-18 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&2&6&-25&19&-18\\& & 4& 20& -10& \color{black}{18} \\ \hline &\color{blue}{2}&\color{blue}{10}&\color{blue}{-5}&\color{blue}{9}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 2x^{4}+6x^{3}-25x^{2}+19x-18 }{ x-2 } = \color{blue}{2x^{3}+10x^{2}-5x+9} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -2 = 0 $ ( $ x = \color{blue}{ 2 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&2&6&-25&19&-18\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 2 }&6&-25&19&-18\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 2 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&2&6&-25&19&-18\\& & \color{blue}{4} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 4 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrr}2&2&\color{orangered}{ 6 }&-25&19&-18\\& & \color{orangered}{4} & & & \\ \hline &2&\color{orangered}{10}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 10 } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&2&6&-25&19&-18\\& & 4& \color{blue}{20} & & \\ \hline &2&\color{blue}{10}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -25 } + \color{orangered}{ 20 } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrr}2&2&6&\color{orangered}{ -25 }&19&-18\\& & 4& \color{orangered}{20} & & \\ \hline &2&10&\color{orangered}{-5}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&2&6&-25&19&-18\\& & 4& 20& \color{blue}{-10} & \\ \hline &2&10&\color{blue}{-5}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 19 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrr}2&2&6&-25&\color{orangered}{ 19 }&-18\\& & 4& 20& \color{orangered}{-10} & \\ \hline &2&10&-5&\color{orangered}{9}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 9 } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&2&6&-25&19&-18\\& & 4& 20& -10& \color{blue}{18} \\ \hline &2&10&-5&\color{blue}{9}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -18 } + \color{orangered}{ 18 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}2&2&6&-25&19&\color{orangered}{ -18 }\\& & 4& 20& -10& \color{orangered}{18} \\ \hline &\color{blue}{2}&\color{blue}{10}&\color{blue}{-5}&\color{blue}{9}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}+10x^{2}-5x+9 } $ with a remainder of $ \color{red}{ 0 } $.