Divide using synthetic division $$ ( -2x^{4}+9x^{3}+5x^{2}+1 ) \div ( x-2 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&-2&9&5&0&1\\& & -4& 10& 30& \color{black}{60} \\ \hline &\color{blue}{-2}&\color{blue}{5}&\color{blue}{15}&\color{blue}{30}&\color{orangered}{61} \end{array} $$The solution is:
$$ \dfrac{ -2x^{4}+9x^{3}+5x^{2}+1 }{ x-2 } = \color{blue}{-2x^{3}+5x^{2}+15x+30} ~+~ \dfrac{ \color{red}{ 61 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&-2&9&5&0&1\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ -2 }&9&5&0&1\\& & & & & \\ \hline &\color{orangered}{-2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&-2&9&5&0&1\\& & \color{blue}{-4} & & & \\ \hline &\color{blue}{-2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrr}2&-2&\color{orangered}{ 9 }&5&0&1\\& & \color{orangered}{-4} & & & \\ \hline &-2&\color{orangered}{5}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 5 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&-2&9&5&0&1\\& & -4& \color{blue}{10} & & \\ \hline &-2&\color{blue}{5}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 10 } = \color{orangered}{ 15 } $
$$ \begin{array}{c|rrrrr}2&-2&9&\color{orangered}{ 5 }&0&1\\& & -4& \color{orangered}{10} & & \\ \hline &-2&5&\color{orangered}{15}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 15 } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&-2&9&5&0&1\\& & -4& 10& \color{blue}{30} & \\ \hline &-2&5&\color{blue}{15}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 30 } = \color{orangered}{ 30 } $
$$ \begin{array}{c|rrrrr}2&-2&9&5&\color{orangered}{ 0 }&1\\& & -4& 10& \color{orangered}{30} & \\ \hline &-2&5&15&\color{orangered}{30}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 30 } = \color{blue}{ 60 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&-2&9&5&0&1\\& & -4& 10& 30& \color{blue}{60} \\ \hline &-2&5&15&\color{blue}{30}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 60 } = \color{orangered}{ 61 } $
$$ \begin{array}{c|rrrrr}2&-2&9&5&0&\color{orangered}{ 1 }\\& & -4& 10& 30& \color{orangered}{60} \\ \hline &\color{blue}{-2}&\color{blue}{5}&\color{blue}{15}&\color{blue}{30}&\color{orangered}{61} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -2x^{3}+5x^{2}+15x+30 } $ with a remainder of $ \color{red}{ 61 } $.