Divide using synthetic division $$ ( 7x^{3}+5x^{2}-7x-2 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&7&5&-7&-2\\& & 14& 38& \color{black}{62} \\ \hline &\color{blue}{7}&\color{blue}{19}&\color{blue}{31}&\color{orangered}{60} \end{array} $$The solution is:
$$ \frac{ 7x^{3}+5x^{2}-7x-2 }{ x-2 } = \color{blue}{7x^{2}+19x+31} ~+~ \frac{ \color{red}{ 60 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{2}&7&5&-7&-2\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 7 }&5&-7&-2\\& & & & \\ \hline &\color{orangered}{7}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 7 } = \color{blue}{ 14 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&7&5&-7&-2\\& & \color{blue}{14} & & \\ \hline &\color{blue}{7}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 14 } = \color{orangered}{ 19 } $
$$ \begin{array}{c|rrrr}2&7&\color{orangered}{ 5 }&-7&-2\\& & \color{orangered}{14} & & \\ \hline &7&\color{orangered}{19}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 19 } = \color{blue}{ 38 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&7&5&-7&-2\\& & 14& \color{blue}{38} & \\ \hline &7&\color{blue}{19}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 38 } = \color{orangered}{ 31 } $
$$ \begin{array}{c|rrrr}2&7&5&\color{orangered}{ -7 }&-2\\& & 14& \color{orangered}{38} & \\ \hline &7&19&\color{orangered}{31}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 31 } = \color{blue}{ 62 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&7&5&-7&-2\\& & 14& 38& \color{blue}{62} \\ \hline &7&19&\color{blue}{31}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 62 } = \color{orangered}{ 60 } $
$$ \begin{array}{c|rrrr}2&7&5&-7&\color{orangered}{ -2 }\\& & 14& 38& \color{orangered}{62} \\ \hline &\color{blue}{7}&\color{blue}{19}&\color{blue}{31}&\color{orangered}{60} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 7x^{2}+19x+31 } $ with a remainder of $ \color{red}{ 60 } $.