Divide using synthetic division $$ ( x^{5}-7x^{4}+4x^{3}+15x^{2}-28 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}2&1&-7&4&15&0&-28\\& & 2& -10& -12& 6& \color{black}{12} \\ \hline &\color{blue}{1}&\color{blue}{-5}&\color{blue}{-6}&\color{blue}{3}&\color{blue}{6}&\color{orangered}{-16} \end{array} $$The solution is:
$$ \frac{ x^{5}-7x^{4}+4x^{3}+15x^{2}-28 }{ x-2 } = \color{blue}{x^{4}-5x^{3}-6x^{2}+3x+6} \color{red}{~-~} \frac{ \color{red}{ 16 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&-7&4&15&0&-28\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}2&\color{orangered}{ 1 }&-7&4&15&0&-28\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 1 } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&-7&4&15&0&-28\\& & \color{blue}{2} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 2 } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrrr}2&1&\color{orangered}{ -7 }&4&15&0&-28\\& & \color{orangered}{2} & & & & \\ \hline &1&\color{orangered}{-5}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&-7&4&15&0&-28\\& & 2& \color{blue}{-10} & & & \\ \hline &1&\color{blue}{-5}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrrr}2&1&-7&\color{orangered}{ 4 }&15&0&-28\\& & 2& \color{orangered}{-10} & & & \\ \hline &1&-5&\color{orangered}{-6}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&-7&4&15&0&-28\\& & 2& -10& \color{blue}{-12} & & \\ \hline &1&-5&\color{blue}{-6}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrrr}2&1&-7&4&\color{orangered}{ 15 }&0&-28\\& & 2& -10& \color{orangered}{-12} & & \\ \hline &1&-5&-6&\color{orangered}{3}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 3 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&-7&4&15&0&-28\\& & 2& -10& -12& \color{blue}{6} & \\ \hline &1&-5&-6&\color{blue}{3}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 6 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrrr}2&1&-7&4&15&\color{orangered}{ 0 }&-28\\& & 2& -10& -12& \color{orangered}{6} & \\ \hline &1&-5&-6&3&\color{orangered}{6}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 6 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&-7&4&15&0&-28\\& & 2& -10& -12& 6& \color{blue}{12} \\ \hline &1&-5&-6&3&\color{blue}{6}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -28 } + \color{orangered}{ 12 } = \color{orangered}{ -16 } $
$$ \begin{array}{c|rrrrrr}2&1&-7&4&15&0&\color{orangered}{ -28 }\\& & 2& -10& -12& 6& \color{orangered}{12} \\ \hline &\color{blue}{1}&\color{blue}{-5}&\color{blue}{-6}&\color{blue}{3}&\color{blue}{6}&\color{orangered}{-16} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}-5x^{3}-6x^{2}+3x+6 } $ with a remainder of $ \color{red}{ -16 } $.