Divide using synthetic division $$ ( 2x^{5}+x^{4}+15x-12 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}2&2&1&0&0&15&-12\\& & 4& 10& 20& 40& \color{black}{110} \\ \hline &\color{blue}{2}&\color{blue}{5}&\color{blue}{10}&\color{blue}{20}&\color{blue}{55}&\color{orangered}{98} \end{array} $$The solution is:
$$ \frac{ 2x^{5}+x^{4}+15x-12 }{ x-2 } = \color{blue}{2x^{4}+5x^{3}+10x^{2}+20x+55} ~+~ \frac{ \color{red}{ 98 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&2&1&0&0&15&-12\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}2&\color{orangered}{ 2 }&1&0&0&15&-12\\& & & & & & \\ \hline &\color{orangered}{2}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&2&1&0&0&15&-12\\& & \color{blue}{4} & & & & \\ \hline &\color{blue}{2}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 4 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrrr}2&2&\color{orangered}{ 1 }&0&0&15&-12\\& & \color{orangered}{4} & & & & \\ \hline &2&\color{orangered}{5}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&2&1&0&0&15&-12\\& & 4& \color{blue}{10} & & & \\ \hline &2&\color{blue}{5}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 10 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrrr}2&2&1&\color{orangered}{ 0 }&0&15&-12\\& & 4& \color{orangered}{10} & & & \\ \hline &2&5&\color{orangered}{10}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 10 } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&2&1&0&0&15&-12\\& & 4& 10& \color{blue}{20} & & \\ \hline &2&5&\color{blue}{10}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 20 } = \color{orangered}{ 20 } $
$$ \begin{array}{c|rrrrrr}2&2&1&0&\color{orangered}{ 0 }&15&-12\\& & 4& 10& \color{orangered}{20} & & \\ \hline &2&5&10&\color{orangered}{20}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 20 } = \color{blue}{ 40 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&2&1&0&0&15&-12\\& & 4& 10& 20& \color{blue}{40} & \\ \hline &2&5&10&\color{blue}{20}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ 40 } = \color{orangered}{ 55 } $
$$ \begin{array}{c|rrrrrr}2&2&1&0&0&\color{orangered}{ 15 }&-12\\& & 4& 10& 20& \color{orangered}{40} & \\ \hline &2&5&10&20&\color{orangered}{55}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 55 } = \color{blue}{ 110 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&2&1&0&0&15&-12\\& & 4& 10& 20& 40& \color{blue}{110} \\ \hline &2&5&10&20&\color{blue}{55}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 110 } = \color{orangered}{ 98 } $
$$ \begin{array}{c|rrrrrr}2&2&1&0&0&15&\color{orangered}{ -12 }\\& & 4& 10& 20& 40& \color{orangered}{110} \\ \hline &\color{blue}{2}&\color{blue}{5}&\color{blue}{10}&\color{blue}{20}&\color{blue}{55}&\color{orangered}{98} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{4}+5x^{3}+10x^{2}+20x+55 } $ with a remainder of $ \color{red}{ 98 } $.