Divide using synthetic division $$ ( 5x^{3}+4x^{2}-6x+7 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&5&4&-6&7\\& & 10& 28& \color{black}{44} \\ \hline &\color{blue}{5}&\color{blue}{14}&\color{blue}{22}&\color{orangered}{51} \end{array} $$The solution is:
$$ \frac{ 5x^{3}+4x^{2}-6x+7 }{ x-2 } = \color{blue}{5x^{2}+14x+22} ~+~ \frac{ \color{red}{ 51 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{2}&5&4&-6&7\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 5 }&4&-6&7\\& & & & \\ \hline &\color{orangered}{5}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 5 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&5&4&-6&7\\& & \color{blue}{10} & & \\ \hline &\color{blue}{5}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 10 } = \color{orangered}{ 14 } $
$$ \begin{array}{c|rrrr}2&5&\color{orangered}{ 4 }&-6&7\\& & \color{orangered}{10} & & \\ \hline &5&\color{orangered}{14}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 14 } = \color{blue}{ 28 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&5&4&-6&7\\& & 10& \color{blue}{28} & \\ \hline &5&\color{blue}{14}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 28 } = \color{orangered}{ 22 } $
$$ \begin{array}{c|rrrr}2&5&4&\color{orangered}{ -6 }&7\\& & 10& \color{orangered}{28} & \\ \hline &5&14&\color{orangered}{22}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 22 } = \color{blue}{ 44 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&5&4&-6&7\\& & 10& 28& \color{blue}{44} \\ \hline &5&14&\color{blue}{22}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 44 } = \color{orangered}{ 51 } $
$$ \begin{array}{c|rrrr}2&5&4&-6&\color{orangered}{ 7 }\\& & 10& 28& \color{orangered}{44} \\ \hline &\color{blue}{5}&\color{blue}{14}&\color{blue}{22}&\color{orangered}{51} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{2}+14x+22 } $ with a remainder of $ \color{red}{ 51 } $.