Divide using synthetic division $$ ( x^{4}+7x^{3}+5x^{2}-4x+15 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&1&7&5&-4&15\\& & 2& 18& 46& \color{black}{84} \\ \hline &\color{blue}{1}&\color{blue}{9}&\color{blue}{23}&\color{blue}{42}&\color{orangered}{99} \end{array} $$The solution is:
$$ \frac{ x^{4}+7x^{3}+5x^{2}-4x+15 }{ x-2 } = \color{blue}{x^{3}+9x^{2}+23x+42} ~+~ \frac{ \color{red}{ 99 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&7&5&-4&15\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 1 }&7&5&-4&15\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&7&5&-4&15\\& & \color{blue}{2} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 2 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrr}2&1&\color{orangered}{ 7 }&5&-4&15\\& & \color{orangered}{2} & & & \\ \hline &1&\color{orangered}{9}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 9 } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&7&5&-4&15\\& & 2& \color{blue}{18} & & \\ \hline &1&\color{blue}{9}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 18 } = \color{orangered}{ 23 } $
$$ \begin{array}{c|rrrrr}2&1&7&\color{orangered}{ 5 }&-4&15\\& & 2& \color{orangered}{18} & & \\ \hline &1&9&\color{orangered}{23}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 23 } = \color{blue}{ 46 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&7&5&-4&15\\& & 2& 18& \color{blue}{46} & \\ \hline &1&9&\color{blue}{23}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 46 } = \color{orangered}{ 42 } $
$$ \begin{array}{c|rrrrr}2&1&7&5&\color{orangered}{ -4 }&15\\& & 2& 18& \color{orangered}{46} & \\ \hline &1&9&23&\color{orangered}{42}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 42 } = \color{blue}{ 84 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&7&5&-4&15\\& & 2& 18& 46& \color{blue}{84} \\ \hline &1&9&23&\color{blue}{42}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ 84 } = \color{orangered}{ 99 } $
$$ \begin{array}{c|rrrrr}2&1&7&5&-4&\color{orangered}{ 15 }\\& & 2& 18& 46& \color{orangered}{84} \\ \hline &\color{blue}{1}&\color{blue}{9}&\color{blue}{23}&\color{blue}{42}&\color{orangered}{99} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+9x^{2}+23x+42 } $ with a remainder of $ \color{red}{ 99 } $.